Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 5th Lesson General Principles of Metallurgy Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 5th Lesson General Principles of Metallurgy

Very Short Answer Questions

Question 1.
What is the role of depressant in froth floatation?
Answer:
By using depressants in the froth floatation process, it is possible to separate a mixture of two sulphide ores.
Eg: In the ore containing ZnS and PbS, the depressant used is NaCN. It prevents ZnS from coming to the froth but allows PbS to come with the froth.

Question 2.
Between C and CO, which is a better reducing agent at 673K.
Answer:

• Out of C and CO, Carbon nonoxide (CO) is a better reducting agent at 673K.
• At 983K (or) above coke(C) is better reducing agent.
• The above observations are form Ellingham diagram.

Question 3.
Name the common elements present in the anode mud in the eletrolytic refining of copper.
Answer:

• The common elements present in the anode mud in eletrolytic refining of copper are less reactive valuable metals, silver (Ag), gold (Au) and platinum (Pt) etc….
• These elements donot lose eletrons at anode and collect under the anode as anode mud.

Question 4.
State the role of silica in the metallurgy of copper.
Answer:
The role of silica in the metallurgy of copper is to àcts as an acidic flux. Silica reacts with the impurities of iron and form slag.
FeO (Gangue) + SiO₂ (flux) → FeSiO₃ (Slag)

Question 5.
Explain “poling. [A.P. Mar.16, 15]
Answer:
When the metals are having the metal oxides as impurities this method is employed. The impure metal is melted and is then covered by carbon powder. Then it is stirred with green wood poles. The reducing gases formed from the green wood and the carbon, reduce the oxides to the metal.
Eg: Cu & Sn metals are refined by this method.

Question 6.
Decribe a method for the refining of nickel.
Answer:
Mond’s process:

• In Mond’s process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl.
  
• Nickel tetra carbornyl is strongly heated to decompose and gives the pure Nickel
  

Question 7.
How is cast iron different from pig iron ?
Answer:

• Cast iron is formed by melting pig iron with scrap iron and coke in presence of blast of hot air. It contains carbon content 3%(approximately). It is extremely hard and brittle.
• Pig iron is formed from blast furnace. It contains carbon 4% (approximately). It contains small quantities of impurities S, P, Si, Mu etc….

Question 8.
What is the difference between a mineral and an ore ?
Answer:
Minerals : The naturally occuring chemical compounds of metal in the earth’s crust are called minerals.
Ore : The mineral from which metal can be extracted economically is called ore.

Question 9.
Why copper matte is put in silica lined converter ?
Answer:
Copper matte conains Cu₂ sand FeS. In this mixture FeS is gangue. For removing the gangue, silica present in the lining of the Bessemer’s converter acts as acidic flux and forms slag.
2FeS (gangue) + 3O₂ → 2FeO + 2SO₂
FeO (gangue) + SiO₂ (flux) → FeSiO₃ (Slag)

Question 10.
What is the role of cryolite in the metallurgy of aluminium? [T.S. Mar. 16, 15]
Answer:
By adding the cryolite to the pure Alumina, the melting point of pure Alumina is lowered (which is very high 2324K) and eletrical conductivity of pure alumina is increased.

Question 11.
How is leaching carried out in the case of low grade copper ores?
Answer:
In case of low grade ores of copper, hydrometallurgy technique is used for extraction. Here leaching process can be done by using acids (or) bacteria, The solution containing Cu⁺² is treated with scrap iron (or) H₂.
Cu⁺²_(aq) + H_2(g) → Cu_(S) + 2H⁺_(aq)

Question 12.
Why is zinc not extracted from zinc oxide through reduction using CO?.
Answer:
Zinc is not extracted from zinc oxide through reduction by using CO.
Explanation:
2Zn + O₂ → 2ZnO, ∆G° = -650 IcI
2C0 + O₂ → 2CO, ∆G° = -450 kJ
2ZnO + 2CO → 2Zn, 2CO₂, ∆G° = 200 kJ
∆G° = Positive indicatis that the reaction is not feasible.

Question 13.
Give the composition of the following alloys. [A.P. & T.S. Mar. 19, 17] [Mar. 14]
a) Brass
b) Bronze
c) German Silver
Answa:
a) Composition of Brass : 60 – 80% Cu, 20 – 40% Zn .
b) Composition of Bronze : 75 – 90% Cu, 10 – 25% Sn
c) Composition of German silver : 50 – 60% Cu, 10 – 30% Ni, 20 – 30% Zn.

Question 14.
Explain the terhts gangue and slag.
Answer:
Gangue : The ore is contaminated with the minerals of earth crust and undesired chemical compounds. These are known as gangue (or) matrix.
Slag: The fused material obtained during the refining (or) smelting of metals by combining the flux with gangue is called slag.
Eg : FeO (gangue) + SiO₂ (flux) → FeSiO₃ (Slag)

Question 15.
How is Ag or Au obtained by leaching from the respective ores ?
Answer:
Ag and Au are leached with a dilute solution of NaCN (or) KCN in presence of 02(air). From the leached solution the metal is obtained through displacement by Zinc. .
4M_(s) + 8eN^–_(aq) + 2H₂O_(aq) + O_2(g) → 4[M(CN)₂]^–_(aq) + 4OH^–_(aq)
2[M(CN)₂ ]^–_(aq) + Zn_(s) → [Zn(CN)₄]^2-_(aq) + 2M_(s)

Question 16.
What are the limitations of Ellingham diagram ?
Answer:
Limitations of Ellingham diagram :

• Ellingham diagram is based only on the thermodynamic concepts. It does not explain the kinetics of the reduction process. The graph simply indicates whether a reacton is possible or not but not the kinetics of the reaction.
• The interpretation of ∆G° is depends on K [ ∆G° = -RTl.nK]
  It is presumed that the reactants and products are in equilibrium. .
  M_xO + A_red ⇌ xM + AO_ox
  This is not always true because the reactant or product may be solid.

Question 17.
Write any two ores with formulae of the following metals :
(a) Aluminium
(b) Zinc
(c) Iron
(d) copper
Answer:
a) Ores of Aluminium : Bauxite – Al₂ O₃. 2H₂O
Cryolite – Na₃AlF₆
b) Ores of Zinc : Zinc blende – Zns
Calamine – ZnCO₃
c) Ores of Iron : Haematite – Fe₂O₃
Magnetite – Fe₃O₄
d) Ores of copper : Copper pyrites – CuFeS₂
Copper glance – Cu₂S.

Question 18.
What is matte ? Give its composition.
Answer:
During the extraction of ‘Cu’ from copper pyrites the product of the blast furnace consists mostly of Cu₂S and a little of FeS. This product is known as “Matte”. It is collected from the outlet at the bottom of the fumance.

Question 19.
What is blister copper ? Why is it so called ? [T.S. Mar. 18]
Answer:
During the extraction of ‘Cu’ from copper pyrites when the matte is charged into a Bessemer converter then cuprous oxide combine with cuprous sulphide and forms Cu metal.
2CU₂O + Cu₂S → 6Cu + SO₂
The ‘Cu’ meted is cooled.
The obtained copper metal is impure and is known as “Blister copper” (98% pure).
The solid fied copper has blistered appearance due to evolution of SO₂ Hence it is called blister copper.

Question 20.
Explain magnetic separation of impurities from an ore.
Answer:
Electro-magnetic method : The method is used if the gangue or the ore particles are magnetic in nature. The finely powdered ore is dropped on a belt moving on two strong electromagnetic rollers. The magnetic and the non-magnetic substances form two separate heaps.

For example, let the ore particles be magnetic and then the non-magnetic material will be the gangue. The non-magnetic material forms a heap nearer to the magnetic roller.
Eg : Haematite and magnetite hae magnetic ore particles. Cassiterite or Tin stone has Wolframite as magnetic impurity.

Question 21.
What is flux ? Give an example.
Answer:
Flux : An outside substance added to ore to lower its melting point is known as flux.

• Flux combines with gangue and forms easily fusible slag.
  gangue + flux → slag
  Eg: FeO (gangue) + SiO₂ (flux) → FeSiO₃ (Slag

Question 22.
Give two uses each of the following metals :
(a) Zinc
(b) Copper
(c) Iron
(d) Aluminium
Answer:
a) Uses of Zinc :

• Zinc is used in large quantities in batteries
• Zinc is used for galvanising iron.
• Zinc is used in manufacturing alloys.
  Eg: Brass, German silver,

b) Uses of copper :

• Copper is used in manufacturing of water pipes and steam pipes.
• Copper is used in making wires used in electrical industry.
• Copper is used in manufacturing of alloys.
  Eg : Brass, Bronze. .

c) Uses of Iron:

• Cast iron is used in casting stoves, railway siéepers- gutter pipers, toys etc.
• It is used in manufacturing of wrought iron and steel.
• Wrought iron is used in making anchors, wires, bolts, chains etc.

d) Uses of Aluminium :

• Aluminium foils are used as wrappers for chacolates.
• Aluminium is fine dust used in paints and lacquers.
• Aluminium is used in photo frames.
• Aluminium is used in extraction of Cr and Mn from their oxides.

Question 23.
Between C and CO, which is a better reducing agent for ZnO ?.
Answer:
Case – I : [Coke as reducing agent]
ZnO + C → Zn + CO, ∆G° become lesser as the T is more then 1120K
Case – II : [CO as reducing agent]
ZnO + CO₂ → Zn + CO, ∆G° becomes lesser when the T is more then 1323K.

• The value of ∆G° is negative for a reaction to occur.
• In the equation (1) ∆G° becomes negative at low temperature- so equation (1) is feasible i. r. C is a better reducing agent for ZnO.

Question 24.
Give the uses of
a) Cast iron
b) Wrought iron
c) Nickel steel
d) Stainless steel
Answer:
a) Uses of Cast iron :

• Cast iron is used for casting stoves, railway sleepers, gutter pipes, toys etc.
• It is used in manufacturing of wrought iron and steel.

b) Uses of wrought iron :

• Wrought iron is used in making anchors, wires.
• Wrought iron is used in making chains and agricultural implements.

c) Uses of Nickel steel:

• Nickel steel is used for making cables, automobiles and aeroplane parts.
• Nickel steel is used for making pendulum, measuring tapes, chrome steel for cutting tools and crashing machines.

d) Uses of stainless steel:

• Stainless steel is used in manufacturing of cycles, automobiles.
• Stainless steel is used in manufacturing of utensils, pens etc.

Question 25.
How is aluminium useful in the extraction of chromium and manganese from their oxides ?
Answer:

• ’AT’ is used as reducing agent.
• 4 By Alumino thermite process Cr, Mn are extracted from their oxides.
• The reactions are highly exothermic.
  Cr₂O₃ + 2Al → 2Cr + Al₂O₃
  3Mn₃O₄ + 8Al → 4Al₂O₃ + 9Mn

Short Answer Questions

Question 1.
Copper can be extracted by hydrometallurgy but not zinc – explain.
Answer:
Copper can be extracted by hydromefallurgy but not zinc.
Explanation:

• E⁰ Value of Zn⁺²/Zn = -0.762V is less than that of E⁰ value of Cu⁺²/Cu = 0.337V .
• From the above data zinc is a stronger reducing agent and can easily displace the Cu⁺² ions present in the complex.
  [Cu(CN)₂] + Zn (Soluble complex → [Zn(CN)₂] + Cu (Precipitate)
• Zinc can be isolated by hydrometallurgy only when stronger reducing agents than zinc are present.
• So zinc cannot be extracted by hydrometallurgy.

Question 2.
Why is the extraction of copper form pyrites more difficult then that from its oxide ore through reduction?
Answer:
The extraction of copper from pyrites is more difficult than that from its oxide ore through reduction.
Explanation:
Pyrites (Cu₂S) Cannot be reduced by carbon(or) hydrogent because the standard free energy of formation (∆G°) of Cu₂S is greater then those of CS₂ and H₂S.

The ∆G° of copper oxide is less than that of CO₂.
∴ The sulphide ore is first converted to oxide by roasting and then reduced.

Question 3.
Explain zone refining.
Answer:
Zone refining:

• Zone refining is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
• A circular mobile heater is fixed at one end of a rod of impure metal.
• The molten zone moves along with the heater moves forward the pure metal crystallises out of the melt and the impurities pass into the adjacent molten zone.
• The above process is repeated several times and the heater is moved in the same direction form one end to the other end. At one end impurities get concentrated. This end is cut off.
• This method is very useful for producing semiconductor grade metals of very high purity. Eg : Ge, Si, B, Ga etc…
  

Question 4.
Write down the chemical reactions taking place in the extraction of zinc from zinc blende.
Answer:
In the extraction of zinc from zinc blende the following chemical reactions taking place.
i) Roasting : The process of strong heating the zinc blende in presence of air is roasting.

ii) Reduction : Zinc oxide obtained by roasting undergo reduction with coke.

Question 5.
Write down the chemical reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
The chemical reactions taking place in different zones in the blast furnace during the extraction of iron are

Question 6.
How is alumina separated from silica in the bauxite ore associated with silica ? Give equations.
Answer:
Serpeck’s process is used for the purification of white Bauxite i.e., Bauxite containing silica as impurity.
In this process bauxite is mixed with coke and is then heated to 2075K in a current of nitrogen. Silica is reduced to silicon and escapes as a vapour.

Aluminium nitride formed in the above step is hydrolysed to form aluminium hydroxide.
AlN + 3H₂O → AZ(OH)₃ ↓ + NH₃ ↑
Al(OH)₃ PPt is washed, dried and then ignited to from purified bauxite.

Question 7.
Giving examples to differentiate roasting and calcination. [T.S. Mar. 19, 18, 16, 15; A.P. Mar. 16, 15]
Answer:
Roasting: Removal of the volatile components of a mineral by heating mineral either alone (or) mixed with some other substances to a high temperature in the presence of air is called Roasting.

• It is applied to the sulphide ores.
• SO₂ gas is producted along with metal oxide.
  Eg : 2ZnS + 3O₂ 2ZnO + 2SO₂ ↑

Calcination: Removal of the volatile components of a mineral by heating in the absence of air is called calcination.

•  It is applied to carbonates and bicarbonates.
• CO₂ gas is produced along with metal oxide.
  Eg: CaCO₃ CaO + CO₂
  ZnCO₃ ZnO + CO₂

Question 8.
The Value of ∆G° for the formation of Cr₂O₃ is – 540kJ mol^-1 and that of Al₂O₃ is – 827kJ mol^-1. Is the reduction of Cr₂O₃ Possible with Al?
Answer:
From the given data the following are the thermo chemical equations
\(\frac{4}{3}\) Cr_(s) + O_2(g) → \(\frac{2}{3}\) Cr₂O_2(s), ∆G° – 540 KJ ……………….. (1)
\(\frac{4}{3}\) Al_(s) + O_2(g) → \(\frac{2}{3}\) Al₂O_3(s), ∆G° – 827 KJ ……………….. (2)
Equation (1) – Equation (2)
\(\frac{2}{3}\) Cr₂O_3(s) + \(\frac{4}{3}\) Al_(s) → \(\frac{2}{3}\) Al₂O_3(s) + \(\frac{4}{3}\) Cr_(s), ∆G° – 287 KJ
∆G° = -ve, So the reaction is feasible
Hence the reduction of Cr₂O₃ is possible with ‘Al’.

Question 9.
What is the role of graphite rod in the electrometallurgy of aluminium ? [A.P. Mar. 16]
Answer:

• In the electrolytic reduction of alumina b_y Hall – Heroult process graphite rod acts as anode.
• At anode, O₂ gas is liberated which reacts with the carbon of anode to form CO₂ gas. So these graphite rods are consumed slowly and need to be replaced from time to time.
  Al₂O₃ ⇌ 2Al⁺³ + 3O^-2
  Cathode : Al⁺³ + 3e^– → Al
  Anode : 3O^-2 → 3[O] + 6e^–
  C_(s) + O_2(g) → CO_2(g)

Question 10.
Outline the principles of refining of metals by the following methods.
(a) Zone refining
(b) Electrolytic refining
(c) Poling
(d) Vapour phase refining.
Answer:
a) Zone refining :

• Zone refining is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. .
• A circular mobile heater is fixed at one end of a rod of impure metal.
• The molten zone moves along with the heater moves forward the pure metal crystallises out of the melt and the impurities pass into the adjacent molten zone.
• The above process is repeated several times and the heater is moved in the same direction form one end to the other end. At one end impurities get concentrated. This end is cut off.
• This method is very useful for producing semiconductor grade metals of very high purity.
  Eg : Ge, Si, B, Ga etc..
  

b) Electrolytic refining : This process is used for less reactive metals like Cu, Ag, AZ, Au etc.

• In this process anode is made by impure metal and a thin strip of pure metal acts as cathode. .
• On electrolysis metal dissolves from anode and pure metal gets deposited at cathode.
  M → Mⁿ⁺ + ne^–
  Impure
  Mⁿ⁺ + ne^– → M (Cathode) Pure metal
  Impurities settle down below anode in the form of anode mud.

c) Poling : When the metals are having the metal oxides as impurities this method is employed. The impure metal is melted and is then covered by carbon powder. Then it is stirred with green wood poles. The reducing gases formed from the green wood and the carbon, reduce the oxides to the metal.
Eg : Cu & Sh metals are refined by this method.

d) Vapour phase refining: In this method the metal is converted into its volatile compound and collected. It is then decomposed to give pure metal. .

1. The metal should form a volatile compound with an available reagent.
2. The volatile compound should be easily decomposable. So the the recovery is easy.
   E.g : Mond’s process :
3. In Mond’s process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl.
   
4. Nickel tetra carbomyl is strongly heated to decompose and gives the pure Nickel
   

Question 11.
Predict the conditions under which Al might be expected to reduce MgO.
Answer:
The Equations for the formation of two oxides are
\(\frac{4}{3}\)Al_(S) + O_2(g) → 2Mg_(S) + \(\frac{2}{3}\) O_2(g) \(\frac{4}{3}\) 2MgO_(S)
From the Ellingham diagram the two curves of these oxides formation intersect each otherat a certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by aluminium metal,
2MgO_(S) + \(\frac{4}{3}\) Al_(S) ⇌ 2Mg_(S) + \(\frac{2}{3}\) Al₂O_3(S)

• From the above information the reduction of MgO by Al metal cannot occur below this temperature (1665 K) instead, Mg can reduce  Al₂O₃ to Al below 1665 K.
• Al-Metal can reduce MgO to Mg above 1665K because ∆G° for Al₂O₃ is less compared to that of MgO.
  

Question 12.
Explain the purification of sulphide ore by froth floatation method. [A.P. Mar.’19, 17, 15; T.S. Mar. 15]
Answer:
Froth floatation method :
This method is used to concentrate sulphide ores.

• In this process a suspension of the powdered ore is made with water.
• A rotating paddle is used to agitate the suspension and air is blown into the suspension in presence of an oil.
• Froth is formed as a result of blown of air, which carries the mineral particles.
• To the above slurry froth collectors and stabilizers are added.
• Collectors like pine oil enhance non-wettability of the mineral particles.
• Froth stabilizers like cresol stabilize the froth.
• The mineral particles wet by oil and gangue particles wet by water.
• The broth is light and is skimmed off. The ore particles are then obtained from the froth. By using depressants in froth floatation process, it is possible to separate a mixture of two suplhide ores.
  Eg : In the ore containing ZnS and PbS, the depressant used is NaCN. It prevents ZnS from coming to the froth but allows PbS to come with the froth.

Question 13.
Explain the process of leaching of alumina from bauxite. [T.S. Mar. 17]
Answer:
The principal ore of aluminium, bauxite, usually contains SiO₂, iron oxides and titanium oxide (TiO₂) as impurities, concentration is carried out by digesting the powdered ore with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. The way. Al₂O₃ is leached out as sodium aluminate (and SiO₂ too as sodium silicate) leaving the other impurities behind :
Al₂O₃ (S) + 2NaOH(aq) + 3H₂O(l) → 2 Na[Al(OH)₄](aq)
The aluminate is alkaline in nature and is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated. At this stage, the solution is seeded with freshly prepared samples of hydrated Al₂O₃ which induces the precipitation of Al₂O₃ xH₂O
2 Na[Al(OH)₄](aq) + CO₂(g) → Al₂O₃ xH₂O(s) + 2NaHCO₃(aq)
The sodium silicate remains in the solution and the insoluble hydrated alumina is filtered, dried and heated to give pure Al₂O₃:

Question 14.
What is Ellingham diagram ? What information can be known from this in the reduction of oxides ?
Answer:
The graphical representation of Gibbs energy which provides a sound basis for considering the choice of reducing agent in the reduction of oxides. This graphical representation is known as Ellingham diagram.

• This diagram helps us in predicting the feasibility of thermal reduction of an ore.
• The Criterion of feasibility of a reaction is that at a given temperature, Gibbs energy of the reaction must be negative.
•  Ellingham diagram normally consists plots of ∆G° vs T for formation of oxides of elements.
• The graph indicates whether a reaction is possible or not, i.e., the tendency of reduction with a reducing agent is indicated with a reducing agent is indicated.
• The reducing agent forms its oxide when the metal oxide is reduced. The role of reducing agent is to make the sum of ∆G° values of the two reactions negative.
• Out of C and CO, Carbon monoxide (CO) is a better reducting agent at 673K.
• At 983K (or) above coke(C) is better reducing agent.
• The above observations are form Ellingham diagram.
  Zinc is not extracted from zinc oxide through reduction by using CO.

Explanation :

• 2Zn + O₂ → 2ZnO, ∆G° =-650 kJ
• 2CO + O₂ → 2CO₂, ∆G° = -450 kJ
• 2ZnO + 2CO → 2Zn, 2CO₂, ∆G° = 200 kJ
  ∆G° = Positive indicatis that the reaction is not feasible.
  The above fact is explained on the basis of Ellingham diagram.
  

Question 15.
How is copper extracted from copper pyrites?
Answer:
Extraction of copper from copper pyrites :
Copper pyrite is the main source of copper metal. Various steps involved in the extraction of copper discussed below.
Step – I:
Concentration of ore by froth floatation process :
The ore is first crushed in ball mills. The finely divided ore is suspended in water. A little pine oil is added and the mixture is vigorously agitated by a current of air. The froth formed carries the ore particles almost completely. The gangue sinks to the bottom of the tank. The froth is separated and about 95% concentrated ore is obtained.

Step -II:
Roasting : To remove the volatile impurities like As (or) Sb, the ore is roasted in a free supply of air. A mixture of sulphides of copper and iron are obtained and these are partially oxidised to respective oxides.
Cu₂S. Fe₂S₃ + O₂ → Cu2S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂

Step -III:
The roasted ore is mixed with a little coke and sand (Silica) and smelted in a blast furnace and fused. A blast of air, necessary for the combustion of coke, is blown through the tuyeres present at the base of the furnace. The oxidation of the sulphides of copper and iron will be completed further. A slag of iron silicate is formed according to the reactions given below :
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → Fe SiO₃ Ferrous Silicate (Slag)
Cu₂O + FeS → Cu₂S + FeO

Step -IV : After smelting the copper ore in blast furnace, the product of the blast furnace consists mostly of Cu₂S and a little of ferrous sulphide. This product is known as “Matte.” It is collected from the outlet at the bottom of the furnace. After then the following processes are carried out for getting the pure copper.

Bessemerization : The matte is charged into a Bessemer converter. A bessemer converter is a pear-shaped furnace. It is made of steel plates. The furnace is given a basic lining with lime or magnesium oxide (obtained from dolomite or magnesite). The converter is held in position by trunnions and can be tilted in any position. A hot blast of air and sand is blown through the tuyeres present near the bottom. Molten metal, the product in the furnace, collects at the bottom of the converter.

Reactions that took place in blast furnace go to completion. Almost all of iron is eliminated as a slag. Cuprous oxide combines with cuprous sulphide and forms Cu metal.
2Cu₂O + Cu₂S → 6Cu + SO₂
The molten metal is cooled in sand moulds. S02 escapes. The impure copper metal is known as “Blister copper” and is about 98% pure.

Step -V:
Refining : The Blister copper is purified by electrolysis. The impure copper metal is made into plates. They are suspended into lead – lined tanks containing Copper (II) Sulphate solution. Thin plates of pure copper serve as cathode. The cathode plates are coated with graphite. On electrolysis, pure copper is deposited at the cathode. The copper obtained is almost 100% pure Cu.

Question 16.
Explain the extraction of Zinc form Zinc blende.
Answer:
“Zinc blende (ZnS)” is an important source for ‘Zn’ metal.
Extraction of Zinc : Zinc blende ore is treated in the following stages.
i) Crushing : The ore is crushed to a fine powder in ball mills.

ii) Concentration of the ore : The ore is concentrated first by gravity process. The crushed ore is washed with a stream of water on a Wilfley’s table. The tables have a corrugated top and are under a rocking motion. Due to this motion, the lighter gangue particles are washed away by the steam. The heavier ore particles settle to the bottom of the table.

The partially concentrated ore is further concentrated by froth floatation process. The ore particles go with the froth.

The concentrated ore is subjected further to electromagnetic separation, if iron oxide is present in the gangue. Iron oxide is magnetic and so forms a heap nearer to the magnetic pole.

iii) Roasting: The concentrated ore thus obtained is roasted in rotary shelf burner which is provided with horizontal shelves and raking arms. The ore is added at the top and zinc oxide is collected from the bottom. The following reactions take place.
2ZnS + 3O₂ → 2ZnO + 2SO₂
ZnS + 2O₂ → ZnSO₄
2ZnSO₄ → 2ZnO + 2SO₂ + O₂.
iv) Reduction : Three methods are in practice for the reduction of the oxide to the metal., The most commonly used method is the Belgian process. In this process the roasted ore is mixed with coal or coke intimately and is taken in fireclay or earthem ware retorts. The retorts are made of fireclay which are bottle shaped tubes. These are closed at one end. The other end is connected to air cooled earthem ware condenser. A large number of retorts are placed in tiers in a large furnace and heated in 1100°C by burning the gas. ‘Prolongs’ made of sheet iron are attached to the condensers. The metal condenses in these earthem ware condensers and the prolongs. The metal powder collected is mixed with some zinc oxide and is known as “zinc dust”. Some of the zinc metal is collected in the fused state. This is solidified in moulds. This metal is called “zinc spelter”.
ZnO + C → Zn + CO
ZnO + CO → Zn + CO₂
Electrolytic refining : Very pure zinc is obtained by electrolysis. The electrolyte is zinc sulphate solution containing a little H₂SO₄. Impure zinc is made the anode and the pure zinc plates are made the cathode.

Question 17.
Explain smelting process in the extraction of copper.
Answer:
The roasted ore is mixed with a little coke and sand (Silica) and smelted in a blast furnace and fused. A blast of air, necessary for the combustion of coke, is blown through the tuyeres present at the base of the furnace. The oxidation of the sulphides of copper and iron will be completed further. A slag of iron silicate is formed according to the reactions given below :
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → Fe SiO₃
Ferrous Silicate (Slag)
Cu₂O + FeS → Cu₂S + FeO

Question 18.
Explain electrometallurgy with an example.
Answer:
Electro metallurgy : Metallurgy involving the use of electric arc furnaces, electrolysis and other electrical operations is called electrometallurgy.
In the reduction of a molten metal salt, electrolysis is used. These methods are based on electro chemical principles.
These are expained by the equation
∆G° = -n FE°
n = no. of electrons
E° = Electrode potential.
Electrolytic Reduction of Alumina : Pure Alumina (Al₂O₃) is a bad conductor of electricity and it has high melting point (2050°C). So it cannot be electrolysed. Alumina is electrolysed by dissolving in fused cryolite to increase the conductivity and small amount of Fluorspar is added to reduce its melting point. Thus the electrolyte is a fused mixture of Alumina, Cryolite and Fluorspar.
Electrolysis is carried out in an iron tank lined inside with graphite (carbon) which functions as cathode. A number of carbon rods (or) copper rods suspended in the electrolyte functions as Anode.

An electric current of 100 amperes at 6 to 7 volts is passed through the electrolyte. Heat produced by the current keeps the mass in fused state at 1175 to 1225K. The following reactions take place in the electrolytic cell under these conditions.
Na₃AlF₆ → 3NaF + AlF₃
Cryolite
4AlF₃ → 4Al³⁺ + 12F^–
At cathode : 4Al³⁺ + 12e^– → 4Al
At anode : 12F^– → 6F₂ + 12e^–
F₂ formed at the anode reacts with alumina and forms Aluminium fluoride.
2Al₂O₃ + 6F₂ → 4AlF₃ + 3O₂
Aluminium, produced at the cathode, sinks to the bottom of the cell. It is removed from time to time through topping hole.

Question 19.
Explain briefly the extraction of aluminium from bauxite.
Answer:
Extraction of Aluminium from Bauxite:
For the purpose of extraction of Al, Bauxite is the best source.
purification of Bauxite : Bauxite containing Fe₂O₃ as impurity is known as red Bauxite.
Bauxite containing SiO₂ as impurity is known as white Bauxite and can be purified by “Serpeck’s Process”. Red Bauxite is purified by Bayer’s process and Hall’s process.
Bayer’s Process : Red bauxite, is roasted and digested in concentrated NaOH at 423 K. Bauxite dissolves in NaOH to form sodium meta aluminate while impurity Fe₂O₃ does not dissolve which can be removed by filtration.
Al₂O₃.2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O
The solution which contains sodium meta aluminate is diluted and crystals of AZ(OH)3, are added which serves as seeding orgent. Sodium meta aluminate undergoes hydrolysis to precipitate Al(OH)₃.
2NaAlO₂ + 4H₂O → 2NaOH + 2Al(OH)₃ ↓
Al(OH)₃ is filtered and ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ Al₂O₃ + 3H₂O
Halls’ Process: Red Bauxite is fused with sodium carbonate to form sodium meta aluminate which is extracted with water. The impurity Fe₂O₃ is filtered out.
Al₂O₃ + Na₂CO₃ → 2NaAlO₂ + CO₂
Into the solution of sodium meta aluminate, CO₂ gas is passed to precipitate Al(OH)₃.

The precipitated Al(OH)₃ is ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ → Al₂O₃ + 3H₂O
Serpeck’s Process: Powdered bauxite is mixed with coke and heated to 2075 K in a current of nitrogen gas. Aluminium Nitride is formed while Si02 is reduced to SiO₂ which escapes out.

Aluminium nitride is hydrolysed to get aluminium hydroxide which on ignition gives anhydrous alumina.
AlN + 3H₂O → Al(OH)₃ ↓ + NH₃ ↑
2Al(OH)₃ 3 Al₂O₃ + 3H₂O

Electrolytic Reduction of Alumina : Pure Alumina (Al₂O₃) is a bad conductor of electricity and it has high melting point (2050°C). So it cannot be electrolysed. Alumina is electrolysed by dissolving in fused cryolite to increase the conductivity and small amount of Fluorspar is added to reduce its melt¬ing point. Thus the electrolyte is a fused mixture of Alumina, Cryolite and Fluorspar.

Electrolysis is carried out in an iron tank lined inside with graphite (carbon) which functions as cathode. A number of carbon rods (or) copper rods suspended in the electrolyte functions as Anode.
Na₃AlF₆ → 3NaF + AlF₃
Cryolite
4AlF₃ → 4Al³⁺ + 12F^–
At cathode : 4Al³⁺ + 12e^– → 4Al
At anode : 12F^– → 6F₂ + 12e^–
F₂ formed at the anode reacts with alumina and forms Aluminium fluoride.
2Al₂O₃ + 6F₂ → 4AlF₃ + 3O₂
Aluminium, produced at the cathode, sinks to the bottom of the cell. It is removed from time to time through topping hole.

Purification of Aluminium : (Hoope’s Process)
The impurities present are Si, Cu, Mn etc.,
The electrolytic cell used for refining of aluminium consists of iron tank lined inside with carbon. This acts as anode. The tank contains three layers of fused masses. The bottom layer contains impure aluminium. Middle layer contains mixture of AlF₃, NaF and BaF₂ saturated with Al₂O₃. Top layer contains pure aluminium and graphite rods kept in it act as cathode.

On passing current aluminium ions from the middle layer are discharged at the cathode as pure aluminium. Equivalent amount of aluminium from the bottom layer passes into middle layer.

Long Answer Questions

Question 1.
The choice of a reducing agent in a particular case depends on thermo-dynamic factor, •lain with two examples.
Answer:
The choice of a reducing agent in a particular case depends on thermodynamic factor. This fact can be explained by considering the following examples.

• Out of C and CO, Carbon nonoxide (CO) is a better reducting agent at 673K.
• At 983K (or) above coke(C) is better reducing agent.
• The above observations are form Ellingham diagram.
• Zinc is not extracted from zinc oxide through reduction using CO.

Explanation :
2Zn + O₂ → 2ZnO, ∆G° = -650 kJ
2CO + O₂ → 2CO₂, ∆G° = -450 kJ
2ZnO + 2CO → 2Zn + 2CO₂, ∆G° = 200 kJ
∆G° = Positive indicatis that the reaction is not feasible.
The extraction of copper from pyrites is more difficult than that from its oxide ore through reduction. ‘ .
Explanation : Pyrites (Cu₂S) Cannot be reduced by carbon (or) hydrogent because the standard free energy of formation (∆G°) of Cu₂S is greater then those of CS₂ and H₂S.

The ∆G° of copper oxide is less than that of CO₂.
∴ The sulphide ore is first converted to oxide by roasting and then reduced. Roasting

The following are the conditions that Al might be expected to reduce MgO. The Equations for the formation of two oxides are
\(\frac{4}{3}\)Al_(S) + O_2(g) → \(\frac{2}{3}\) Al₂O_3(S)
2Mg_(S) + O_2(g) → 2MgO_(S)
From the Ellingham diagram the two curves of these oxides formation intersect each other at a certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by aluminium metal,
2MgO_(S) + \(\frac{4}{3}\)Al_(S) → 2Mg_(S) + \(\frac{2}{3}\) Al₂O_3(S)

• From the above information the reduction of MgO by Al metal cannot occur below this temperature (1665 K) instead, Mg can reduce Al₂O₃ to Al below 1665 K.
• Al-Metal can reduce MgO to Mg above 1665 K because ∆G° for Al₂O₃ is less compared to that of MgO.
  

Question 2.
Discuss the extraction of Zinc from Zinc blend
Answer:
“Zinc blende (ZnS)” is an important source for ‘Zn’metal.
Extraction of’Zinc : Zinc blende ore is treated in the following states.
i) Crushing : The ore is crushed to a fine powder in ball mills.
ii) Concentration of the ore : The ore is concentrated first by gravity process. The crushed ore is washed with a stream of water on a Wilfley’s table. The tables have a corrugated top and are under a rocking motion. Due to this motion, the lighter gangue particles are washed away by the steam. The heavier ore particles settle to the bottom of the table.

The partially concentrated ore is further concentrated by froth floatation process. The ore particles go with the froth.

The concentrated ore is subjected further to electromagnetic separatioh, if iron oxide is present in the gangue. Iron oxide is magnetic and so forms a heap nearer to the magnetic pole.

iii) Roasting: The concentrated ore thus obtained is roasted in rotary shelf burner which is provided with horizontal shelves and raking arms. The ore is added at the top and zinc oxide is collected from the bottom. The following reactions take place.
2ZnS + 3O₂ → 2ZnO + 2SO₂
ZnS + 2O₂ → ZnSO₄
2ZnSO₄ → 2ZnO + 2SO₂ + O₂
iv) Reduction : Three methods are in practice for the reduction of the oxide to the metal. The most commonly used method is the Belgian process. In this process the roasted ore is mixed with coal or coke intimately and is taken in fireclay or earthern ware retorts. The retorts are made of fireclay which are bottle shaped tubes. These are closed at one end. The other end is connected to air cooled earthern ware condenser. A large number of retorts are placed in tiers in a large furnace and heated in 1I00°C by burning the gas. ‘Prolongs’ made of sheet iron are attached to the condensers. The metal condenses in these earthern ware condensers and the prolongs. The metal powder collected is mixed with some zinc oxide and is known as “zinc dust”. Some of the zinc metal is collected in the fused state. This is solidified in moulds. This metal is called “zinc spelter”.
ZnO + C → Zn + CO
ZnO + CO → Zn + CO₂
Electrolytic refining : Very pure zinc is obtained by electrolysis. The electrolyte is zinc sulphate solution containing a little H₂SO₄. Impure zinc is made the anode and the pure zinc plates are made the cathode.

Question 3.
Explain the reactions occuring in the blast furnace in the extraction of iron.
Answer:
In the Blast furnace, reduction of iron oxides takes place in different temperature ranges. Hot air is blown from the bottom of the furnace and coke is burnt to give temperature upto about 2200K in the lower portion itself. The burning of coke therefore supplies most of the heat required in the process. The CO and heat moves to upper part of the furnace. In upper part, the temperature is 1o%er and the iron oxides (Fe₂O₃ and Fe₃O₄) coming from the top are reduced in steps to FeO. Thus, he reduction reactions taking place in the lower temperature range and in the higher temperature range, depend on the points of corresponding intersections in the ∆_rG° vs T plots.

These reactions can be summarised as follows:
At 500 – 800 K (lower temperature range in the blast furnace)
3 Fe₂O₃ + CO → 2 Fe₃O₄ + CO₂
Fe₃O₄ + 4 CO → 3 Fe + 4 CO₂
Fe₂O₃ + CO → 2 FeO + CO₂
At 900 – 1500 K (higher temperature range in the blast furnace)
C + CO₂ → 2 CO
FeO + CO → Fe + CO₂
Lime stone is also decomposed to CaO which removes silicate impurity of the ore as CaSiO₃ slag. The slag is in molten state and separates out from iron.

The iron obtained from Blast furnace contains about 4% carbon and many impurities in smaller amount (e.g., S, P, Si, Mn). This is known as pig iron. Cast iron is different from and is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.
Fe₂O₃ 3 C → 2 Fe + 3 CO

Question 4.
Discuss the extraction of copper from copper pyrites.
Answer:
Extraction of copper from copper pyrites:
Copper pyrite is the main source of copper metal. Various steps involved in the extraction of copper discussed below.
Step – I:
Concentration of ore by froth floatation process:
The ore is first crushed in ball mills. The finely divided ore is suspended in water. A little pine oil is added and the mixture is vigorously agitated by a current of air. The froth formed carries the ore particles almost completely. The gangue sinks to the bottom of the tank. The froth is separated and about 95% concentrated ore is obtained.

Step – II: Roasting :
To remove the volatile impurities like As (or) Sb, the ore is roasted in a free supply of air. A mixture of sulphides of copper and iron are obtained and these are partially oxidised to respective oxides.
Cu₂S. Fe₂S₃ + O₂ → Cu₂S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂

Step – III :
The roasted ore is mixed with a little coke and sand (Silica) and smelted in a blast furnace and fused. A blast of air, necessary for the combustion of coke, is blown through the tuyeres present at the base of the furnace. The oxidation of the sulphides of copper and iron will be completed further. A slag of iron silicate is formed according to the reactions given below :
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → Fe SiO₃
Ferrous Silicate (Slag)
Cu₂O + FeS → Cu₂S + FeO
Step -IV:
After smelting the copper ore in blast furnace, the product of the blast furnace consists mostly of Cu2S and a little of ferrous sulphide. This product is known as “Matte.” It is collected from the outlet at the bottom of the furnace . After then the following processes are carried out for getting the pure copper.

1. Bessemerization: The matte is charged into a Bessemer converter. A bessemer converter is a pear-shaped furnace. It is made of steel plates. The furnace is given a basic lining with lime or magnesium oxide (obtained from dolomite or magnesite). The converter is held in position by trunnions and can be tilted in any position. A hot blast of air and sand is blown through the tuyeres present near the bottom. Molten metal, the product in the furnace, collects at the bottom of the converter.

Reactions that took place in blast furnace go to completion. Almost all of iron is eliminated as a slag. Cuprous oxide combines with cuprous sulphide and forms Cu metal.
2Cu₂O + Cu₂S → 6Cu + SO₂

The molten metal is cooled in sand moulds. S02 escapes. The impure copper metal is known as “Blister copper” and is about 98% pure.

Step -V:
Refining : The Blister copper is purified by electrolysis. The impure copper metal is made into plates. They are suspended into lead – lined tanks containing Copper (II) Sulphate solution. Thin plates of pure copper serve as cathode. The cathode plates are coated with graphite. On electrolysis, pure copper is deposited at the cathode. The copper obtained is almost 100% pure Cu.

Question 5.
Explain the various steps involved in the extraction of almninium from bauxite.
Answer:
Extraction of Aluminium from Bauxite:
For the purpose of extraction of Al, Bauxite is the best source.
purification of Bauxite : Bauxite containing Fe₂O₃ as impurity is known as red Bauxite.
Bauxite containing SiO₂ as impurity is known as white Bauxite and can be purified by “Serpeck’s Process”. Red Bauxite is purified by Bayer’s process and Hall’s process.
Bayer’s Process : Red bauxite, is roasted and digested in concentrated NaOH at 423 K. Bauxite dissolves in NaOH to form sodium meta aluminate while impurity Fe₂O₃ does not dissolve which can be removed by filtration.
Al₂O₃.2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O
The solution which contains sodium meta aluminate is diluted and crystals of AZ(OH)3, are added which serves as seeding orgent. Sodium meta aluminate undergoes hydrolysis to precipitate Al(OH)₃.
2NaAlO₂ + 4H₂O → 2NaOH + 2Al(OH)₃ ↓
Al(OH)₃ is filtered and ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ Al₂O₃ + 3H₂O
Halls’ Process: Red Bauxite is fused with sodium carbonate to form sodium meta aluminate which is extracted with water. The impurity Fe₂O₃ is filtered out.
Al₂O₃ + Na₂CO₃ → 2NaAlO₂ + CO₂
Into the solution of sodium meta aluminate, CO₂ gas is passed to precipitate Al(OH)₃.

The precipitated Al(OH)₃ is ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ → Al₂O₃ + 3H₂O
Serpeck’s Process: Powdered bauxite is mixed with coke and heated to 2075 K in a current of nitrogen gas. Aluminium Nitride is formed while Si02 is reduced to SiO₂ which escapes out.

Aluminium nitride is hydrolysed to get aluminium hydroxide which on ignition gives anhydrous alumina.
AlN + 3H₂O → Al(OH)₃ ↓ + NH₃ ↑
2Al(OH)₃ 3 Al₂O₃ + 3H₂O

Electrolytic Reduction of Alumina : Pure Alumina (Al₂O₃) is a bad conductor of electricity and it has high melting point (2050°C). So it cannot be electrolysed. Alumina is electrolysed by dissolving in fused cryolite to increase the conductivity and small amount of Fluorspar is added to reduce its melt¬ing point. Thus the electrolyte is a fused mixture of Alumina, Cryolite and Fluorspar.

Electrolysis is carried out in an iron tank lined inside with graphite (carbon) which functions as cathode. A number of carbon rods (or) copper rods suspended in the electrolyte functions as Anode.
Na₃AlF₆ → 3NaF + AlF₃
Cryolite
4AlF₃ → 4Al³⁺ + 12F^–
At cathode : 4Al³⁺ + 12e^– → 4Al
At anode : 12F^– → 6F₂ + 12e^–
F₂ formed at the anode reacts with alumina and forms Aluminium fluoride.
2Al₂O₃ + 6F₂ → 4AlF₃ + 3O₂
Aluminium, produced at the cathode, sinks to the bottom of the cell. It is removed from time to time through topping hole.

Purification of Aluminium : (Hoope’s Process)
The impurities present are Si, Cu, Mn etc.,
The electrolytic cell used for refining of aluminium consists of iron tank lined inside with carbon. This acts as anode. The tank contains three layers of fused masses. The bottom layer contains impure aluminium. Middle layer contains mixture of AlF₃, NaF and BaF₂ saturated with Al₂O₃. Top layer contains pure aluminium and graphite rods kept in it act as cathode.

On passing current aluminium ions from the middle layer are discharged at the cathode as pure aluminium. Equivalent amount of aluminium from the bottom layer passes into middle layer.

Textual Examples

Question 1.
Suggest a condition under which magnesium could reduce alumina.
Answer:
The two equations are :
a) \(\frac{4}{3}\) Al + O₂ → \(\frac{2}{3}\) Al₂O₃
b) 2 Mg + O₂ → 2MgO
At the point of intersection of the Al₂O₃ and MgO curves (marked. “A” in Ellingham diagram), the ∆G^⊖ becomes ZERO for the reaction :
\(\frac{2}{3}\) Al₂O₃ + 2Mg → 2MgO + \(\frac{4}{3}\) Al
Below that point magnesium can reduce alumina.

Question 2.
Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why ?
Answer:
Temperatures below the point of intersection of Al₂O₃ and MgO curves, magnesium can reduce alumina. But the process will be uneconomical.

Question 3.
Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction ?
Solution:
The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change (∆S) of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of ∆G^⊖becomes more on negative side and the reduction becomes easier.

Question 4.
At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why ?
Solution:
Zinc being above iron in the electrochemical series (more reactive metal is zinc), the reduction will be faster in case zinc scraps are used. But zinc is costlier metal than iron. So using iron scraps will be advisable and advantageous.

Intext Questions

Question 1.
Which of the ores mentioned in Table can be concentrated by magnetic separation method ?
Answer:
Ores in which one of the components (either the impurity or the actual ore) is magnetic can be concentrated.
E.g.: ores containing iron (haematite, magnetite, siderite and iron pyrites).

Question 2.
What is the significance of leaching in the extraction of aluminium ?
Answer:
Leaching is significant as it helps in removing the impurities like SiO₂, Fe₂O₃ etc. from the bauxite ore.

Question 3.
The reaction, Cr₂O₃ + 2 Al → Al₂O₃ + 2 Cr (∆G^⊖ = -421 kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?
Answer:
Certain amount of activation energy is essential even for such reactions which are thermodynamically feasible, therefore heating is required.

Question 4.
Is it true that under certain conditions Mg can reduce Al₂O₃ and Al can reduce MgO ? What are those conditions ?
Answer:
Yes, below 1350°C Mg can reduce Al₂O₃ and above 1350°C, Al can reduce MgO. This can be inferred from ∆G^⊖ vs T plots.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 9 Probability to solve questions creatively.

Intermediate 2nd Year Maths 2A Probability Formulas

→ An experiment that can be repeated any number of times under essentially identical conditions and associated with a set of known results is called a random experiment or trial if the result of any single repetition of the experiment is certain and is any one of the associated set.

→ A combination of elementary events in a trial is called an event.

→ The list of all elementary events in a trail is called list of exhaustive events.

→ Elementary events are said to be equally likely if they have the same chance of happening.

→ If there are n exhaustive equally likely elementary events in a trail and m of them are
favourable to an event A, then \(\frac{m}{n}\) is called the probability of A. It is denoted by P(A).

→ P(A) = \(\frac{\text { Number of favourable cases (outcomes) with respect } A}{\text { Number of all cases (outcomes) of the experiment }}\)

→ The set of all possible outcomes (results) in a trail is called sample space for the trail. It is denoted by ‘S’. The elements of S are called sample points.

→ Let S be a sample space of a random experiment. Every subset of S is called an event.

→ Let S be a sample space. The event Φ is called impossible event and the event S is called certain event in S.

→ Two events A, B in a sample space S are said to be disjoint or mutually exclusive if A ∩ B = Φ

→ Two events A, B in a sample space S are said to be exhaustive if A ∪ B = S.

→ Two events A, B in a sample space S are said to be complementary if A ∪ B = S, A ∩ B = Φ. The complement B of A is denoted by Ā (or) A^c.

→ Let S be a finite sample space. A real valued function P: p(s) → R is said to be a probability function on S if (i) P(A) ≥ 0 ∀ A ∈ p(s) (ii) p(s) = 1

→ A, B, ∈ p(s), A ∩ B = Φ ⇒ P (A ∪ B) = P(A) + P(B). Then P is called probability function and for each A ∈ p(s), P(A) is called the probability of A.

→ If A is an event in a sample space S, then 0 ≤ P(A) ≤ 1.

→ If A is an event in a sample space S, then the ratio P(A) : P̄(Ā) is called the odds favour to A and P̄(A): P(A) is called the odds against to A.

→ If A, B are two events in a sample space S, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

→ If A, B are two events in a sample space, then P(B – A) = P(B) – P(A ∩ B) and P(A – B) = P(A) – P(A ∩ B) .

→ If A, B, C are three events in a sample space S, then
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C).

→ If A, B are two events in a sample space then the event of happening B after the event A happening is called conditional event It is denoted by B/A.

→ If A, B are two events in a sample space S and P(A) ≠ 0, then the probability of B after the event A has occured is called conditional probability of B given A. It is denoted by P\(\left(\frac{B}{A}\right)\)

→ If A, B are two events in a sample space S such that P(A) ≠ o then \(P\left(\frac{B}{A}\right)=\frac{n(A \cap B)}{n(A)}\)

→ Let A, B be two events in a sample space S such that P(A) ≠ 0, P(B) ≠ 0, then

• \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
• \(P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}\)

→ Multiplication theorem on Probability: If A and B are two events of a sample space 5 and P(A) > 0, P(B) > 0 then P(A n B) = P(A), P(B/A) = P(B). P(A/B).

→ Two events A and B are said to be independent if P(A ∩ B) = P(A). P(B). Otherwise A, B are said to be dependent.

→ Bayes’ theorem : Suppose E₁, E₂ ……… E_n are mutually exclusive and exhaustive events of a Random experiment with P(E_i) > 0 for ī = 7, 2, ……… n in a random experiment then we have

\(p\left(\frac{E_{k}}{A}\right)\) = \(\frac{P\left(E_{k}\right) P\left(\frac{A}{E_{k}}\right)}{\sum_{i=1}^{n} P\left(E_{i}\right) P\left(\frac{A}{E_{i}}\right)}\) for k = 1, 2 …… n.

Random Experiment:
If the result of an experiment is not certain and is any one of the several possible outcomes, then the experiment is called Random experiment.

Sample space:
The set of all possible outcomes of an experiment is called the sample space whenever the experiment is conducted and is denoted by S.

Event:
Any subset of the sample space ‘S’ is called an Event.

Equally likely Events:
A set of events is said to be equally likely if there is no reason to expect one of them in preference to the others.

Exhaustive Events:
A set of events is said to be exhaustive of the performance of the experiment always results in the occurrence of at least one of them.

Mutually Exclusive Events:
A set of events is said to the mutually exclusive if happening of one of them prevents the happening of any of the remaining events.

Classical Definition of Probability:
If there are n mutually exclusive equally likely elementary events of an experiment and m of them are favourable to an event A then the probability of A denoted by P(A) is defined as min.

Axiomatic Approach to Probability:
Let S be finite sample space. A real valued function P from power set of S into R is called probability function if
P(A) ≥ 0 ∀ A ⊆ S
P(S) = 1, P(Φ) = 0;
(3) P(A ∪ B) = P(A) + P(B) if A ∩ B = Φ. Here the image of A w.r.t. P denoted by P(A) is called probability of A.

Note:

• P(A) + P(A̅) = 1
• If A₁ ⊆ A₂, then P(A₁) < P(A₂) where A₁, A₂ are any two events.

Odds in favour and odds against an Event:
Suppose A is any Event of an experiment. The odds in favour of Event A is P(A̅) : P(A). The odds against of A is P(A̅) : P(A).

Addition theorem on Probability:
If A, B are any two events in a sample space S, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B).
If A and B are exclusive events

• P(A ∪ B) = P(A) + P(B)
• P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C).

Conditional Probability:
If A and B are two events in sample space and P(A) ≠ 0. The probability of B after the event A has occurred is called the conditional probability of B given A and is denoted by P(B/A).
P(B/A) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\)
Similarly
n(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)

Independent Events:
The events A and B of an experiment are said to be independent if occurrence of A cannot influence the happening of the event B.
i.e. A, B are independent if P(A/B) = P(A) or P(B/A) = P(B).
i.e. P(A ∩ B) = P(A) . P(B).

Multiplication Theorem:
If A and B are any two events in S then
P(A ∩ B) = P(A) P(B/A) if P(A)≠ 0.
P (B) P(A/B) if P(B) ≠ 0.
The events A and B are independent if
P(A ∩ B) = P(A) P(B).
A set of events A₁, A₂, A₃ … A_n are said to be pair wise independent if
P(A_i n A_j) = P(A_i) P(A_j) for all i ≠ J.

Theorem:
Addition Theorem on Probability. If A, B are two events in a sample space S
Then P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Proof:
FRom the figure (Venn diagram) it can be observed that
(B – A) ∪ (A ∩ B) = B, (B – A) ∩ (A ∩ B) = Φ

∴ P(B) = P[(B – A) ∪ (A ∩ B)]
= P(B – A) + P(A ∩ B)
⇒ P(B – A) = P(B) – P(A ∩ B) ………(1)

Again from the figure, it can be observed that
A ∪ (B – A) = A ∪ B, A ∩ (B – A) = Φ
∴ P(A ∪ B) = P[A ∪ (B – A)]
= P(A) + P(B – A)
= P(A) + P(B) – P(A ∩ B) since from (1)
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Theorem:
Multiplication Theorem on Probability.
Let A, B be two events in a sample space S such that P(A) ≠ 0, P(B) ≠ 0, then
i) P(A ∩ B) = P(A)P\(\left(\frac{B}{A}\right)\)
ii) P(A ∩ B) = P(B)P\(\left(\frac{A}{B}\right)\)
Proof:
Let S be the sample space associated with the random experiment. Let A, B be two events of S show that P(A) ≠ 0 and P(B) ≠ 0. Then by def. of confidential probability.
P\(\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}\)
∴ P(B ∩ A) = P(A)P\(\left(\frac{B}{A}\right)\)
Again, ∵P(B) ≠ 0
\(\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
∴ P(A ∩ B) = P(B) . P\(\left(\frac{\mathrm{A}}{\mathrm{B}}\right)\)
∴ P(A ∩ B) = P(A) . P\(\left(\frac{B}{A}\right)\) = P(B).P\(\left(\frac{\mathrm{A}}{\mathrm{B}}\right)\)

Baye’s Theorem or Inverse probability Theorem
Statement:
If A₁, A₂, … and A_n are ‘n’ mutually exclusive and exhaustive events of a random experiment associated with sample space S such that P(A_i) > 0 and E is any event which takes place in conjunction with any one of A_i then
P(A_k/E) = \(\frac{P\left(A_{k}\right) P\left(E / A_{k}\right)}{\sum_{i=1}^{n} P\left(A_{i}\right) P\left(E / A_{i}\right)}\), for any k = 1, 2, ………. n;
Proof:
Since A₁, A₂, … and A_n are mutually exclusive and exhaustive in sample space S, we have A_i ∩ A_j = for i ≠ j, 1 ≤ i, j ≤ n and A₁ ∪ A₂ ∪…. ∪ A_n = S.

Since E is any event which takes place in conjunction with any one of A_i, we have
E = (A₁ ∩ E) ∪ (A₂ ∩ E) ∪ …………….. ∪(A_n ∩ E).

We know that A₁, A₂, ……… A_n are mutually exclusive, their subsets (A₁ ∩ E), (A₂ ∩ E) , … are also
mutually exclusive.

Now P(E) = P(E ∩ A₁) + P(E ∩ A₂) + ………. + P(E ∩ A_n) (By axiom of additively)
= P(A₁)P(E/A₁) + P(A₂)P(E/A₂) + ………… + P(A_n)P(E/A_n)
(By multiplication theorem of probability)
= \(\sum_{i=1}^{n}\)P(A_i)P(E/A_i) …………..(1)

By definition of conditional probability
P(A_k/E) = \(\frac{P\left(A_{k} \cap E\right)}{P(E)}\) for
= \(\frac{P\left(A_{k}\right) P\left(E / A_{k}\right)}{P(E)}\)
(By multiplication theorem)
= \(\frac{P\left(A_{k}\right) P\left(E / A_{k}\right)}{\sum_{i=1}^{n} P\left(A_{I}\right) p\left(E / A_{i}\right)}\) from (1)
Hence the theorem

Andhra Pradesh BIEAP AP Inter 2nd Year Commerce Study Material 1st Lesson Entrepreneurship Textbook Questions and Answers.

AP Inter 2nd Year Commerce Study Material 1st Lesson Entrepreneurship

Essay Answer Questions

Question 1.
Explain the characteristics of entrepreneurs.
Answer:
The following are some of the important characteristics that every successful entrepreneur must possess.
1) Innovation:
Innovation is an important characteristic of an entrepreneur in modem business. The entrepreneur makes arrangements for introducing innovations which help in increasing production on the one hand and reducing costs on the other.

2) Risk-taking :
Risk-taking is another feature of an entrepreneur. He has to pay to all the other factors of production in advance. There are chances that he may be rewarded with a handsome profit or he may suffer a heavy loss. Therefore, risk bearing is the final responsibility of an entrepreneur.

3) Decision making :
An entrepreneur has to take decisions with regard to the establishment of business its management and coordination of various resources. Every activity of the business requires decision-making. The entrepreneur has to take decisions every day which have an impact on the working of the enterprise.

4) Leadership :
An entrepreneur has to be a leader because he is such a person who organise direct, control the functions of the organisation. His personality will influence the working of his subordinates who can praise and appreciate him. An entrepreneur in his role as a leader, not only guide and counsels his persons but he motivates them to achieve goals quickly and efficiently.

5) Organisation of production :
Art entrepreneur procures various factors of production like land, labour, capital, raw materials etc. For manufacturing a product or brining out a service. He assess the viability of different production process and selects one which is most suitable.

6) Planning :
An entrepreneur is the person who plans each and everything in the business. A planning is a process which involves thinking before doing an entrepreneur decides the following : What to do? When to do? How to do? Who do a particular task?

7) Work hard :
Willingness to work hard distinguishes a successful entrepreneur from an unsuccessful one. Most of the successful entrepreneurs work hard endlessly, especially in the beginning and the same becomes their whole life.

8) Desire for high achievement and highly optimistic :
The entrepreneurs should have strong desire to achieve high goals in business. The successful entrepreneurs are not disturbed by present problems and are optimistic for future that situation will become favourable to business in future.

9) Independence :
The successful entrepreneurs do not like to be guided by others and like to be independent in the matters of their business.

10) Foresight:
The entrepreneurs should have a good foresight i.e., they visualize the likely changes take place in market, consumer attitude, technological developments etc., and take timely action accordingly.

Question 2.
Explain the functions of entrepreneurs. [A.P. Mar. 17]
Answer:
An entrepreneur performs all the functions necessary right from the generation of an idea and upto setting up an enterprise. The following are the functions of entrepreneur.

1) Formation of new producing organisation :
The function of an entrepreneur is to rationally combine the factors of production into a new producing organisation.

2) Decision making :
An entrepreneur as a decision maker takes various decisions regarding ascertaining the objective of the enterprise, sources of finance, product mix, pricing policies, promotion strategies, appropriate technology or new equipment.

3) Innovation:
Innovation is the main function of an entrepreneur. Innovation means doing new things or doing of things that are already being done in a new way. An entrepreneur puts science and technology to economic use. Innovative entrepreneurs are essential for rapid industrialisation and economic development.

4) Management:
An entrepreneur performs managerial functions such formulation of production plans, providing raw materials, physical facilities, production facilities, organizing and managing sales.

5) Risk bearing :
An entrepreneur undertakes the responsibility for loss that may arise due to unforeseen contingencies in future. He guarantees interest to creditors, wages to labour and rent to land holders.

6) Supervision, control and direction :
J.S.Mill mentions superintendence, control and direction as entrepreneurial functions. Supervision involves assembling the means, turnout maximum output at minimum cost and supervise the work. The entrepreneur has to regulate and control the flow of goods, the use of finance and machinery. He also has to control and direct the activities of the employees.

7) Planning:
Planning is the first step in the direction of setting up of an enterprise. The entrepreneur prepares a scheme of the proposed project in a formal systematic approach. The authorities, if satisfied fully with the requirements, grants legal sanction of the project.

According to Kilby, an entrepreneur performs the following major four functions.

1. Exchange functions
2. Administrative functions
3. Management and control functions
4. Technological functions

Question 3.
Explain the types of entrepreneurs.
Answer:
Entrepreneurs can be found among various sections of the society viz., farmers, artisoris, workers etc. In a study of American Agriculture, Danhof has classified entrepreneurs into four categories. They are :
a) Innovating entrepreneurs
b) Adoptive or initiative entrepreneurs
c) Fabian entrepreneurs and
d) Drone entrepreneurs.

a) Innovating entrepreneurs :
Schumpeter’s entrepreneur was of this type. He introduces new products, new methods of production and open new markets. The entrepreneurs are aggressive in nature. Innovating entrepreneurs experiments and converts attractive possibilities into practice.

b) Adaptive or initiative entrepreneur:
Entrepreneur of this type are found in under-developed countries. This type of entrepreneurs instead of innovating new things, they just adopt the successful innovations innovated by others. However, some of the innovations made by others, may not suit the needs of underdeveloped countries. In such cases, the initiative innovators may make some changes in the innovations made by innovative entrepreneur so as to suit their requirements.

c) Fabian entrepreneur:
These entrepreneurs neither fall in innovative entrepreneur category nor in adoptive entrepreneur category. These are very cautions people. These entrepreneurs are regid and fundamental in approach. They follow the foot steps of their successors. They are shy to introduce new methods and ideas. Fabian entrepreneurs are no risk takers.

d) Drone entrepreneur :
Drone entrepreneurs are lazy in adopting new methods, but Drone entrepreneurs are more rigid than Fabian entrepreneurs. They resist changes. They are laggards. They may close down their business but they dont adopt for changes. Drone entrepreneurs refuse to adopt changes.

Question 4.
Explain the relation between entrepreneur and entrepreneurship. [A.P. Mar. 17]
Answer:
Entrepreneur is the person, entrepreneurship is the process and enterprise is the creation of the person and output of the process. Though the term entrepreneur is often used interchangeably with entrepreneurship, yet they are conceptually different. The relationship between the two is just like the two sides of the same coin.

The following points highlights the relationship between entrepreneur and entre-preneurship.

1. Entrepreneur is a person : Entrepreneurship is a process.
2. Entrepreneur is organiser : Entrepreneurship is the organisation.
3. Entrepreneur is innovator : Entrepreneurship is innovation.
4. Entrepreneur is risk bearer : Entrepreneurship is risk bearing.
5. Entrepreneur is motivator : Entrepreneurship is motivation.
6. Entrepreneur is creator : Entrepreneurship is the creation.
7. Entrepreneur is visualiser : Entrepreneurship is vision.
8. Entrepreneur is the leader : Entrepreneurship is leadership.
9. Entrepreneur is imitator : Entrepreneurship is imitation.

Question 5.
Explain the role of entrepreneurship in economic development.
Answer:
The role of entrepreneurship in economic development varies from economy to economy depending upon its material resources, industrial climate and the responsiveness of the political system to the entrepreneurial functions. India is a developing economy aims to decentralise industries to mitigate the regional imbalances in levels of economic development.

Small-scale entrepreneurship in such industrial structure plays an important role to achieve balanced regional development. It is clearly believed that small scale industries provide immediate large scale employment, ensure a more equitable distribution of national income and facilitate an effective resource mobilisation of capital and skills which might otherwise remain unutilised.

Role of entrepreneurship in economic development:

1. Entrepreneurship promotes capital formation by mobilising the idle savings of the public.
2. It provides large scale employment and helps to reduce unemployment problem.
3. Entrepreneurship promotes balanced regional development.
4. It helps to reduce concentration of economic power.
5. Entrepreneurship stimulates the equitable distribution of wealth, income and even political power in the interest of the country.
6. Entrepreneurship encourages effective resource mobilisation of capital and skill which might otherwise remain unutilised and idle.
7. It also induces backward and forward linkages which stimulate the process of economic development.
8. Entrepreneurship also promote country’s export trade. It is an important ingradient to economic development.

Question 6.
Explain the opportunities for entrepreneurship in Andhra Pradesh.
Answer:
Andhra Pradesh, with its rich and abundant natural resources base as well as diverse agricultural and forest wealth provides tremendous opportunities for entrepreneurs in the state. The various opportunities for entrepreneurs in A.P. are as follows.

1) Information technology:
Government of A.P. has declared Information Technology (IT) industry as an essential service under the Essential Services Maintenance Act and the industry have been exempted from power cuts. It aims to transform the state into a knowledge society and make available the benefits of IT to all citizens, especially those in rural areas. People with innovative ideas would be encouraged to set up their own IT and IT enabled services units in the state.

2) Automobiles :
A broad base of automotive equipment manufacturers and a large pool of highly trained, skilled and disciplined man power make Andhra Pradesh the desired location for automobile industries. There are more than 100 automotive component manufacturing companies in the state. The industry is highly potential and there is a plenty of demand for automobile components, hence the entrepreneurs in the state may grab the opportunities.

3) Drugs and pharmaceuticals:
The state offers excellent opportunities for the growth of pharmaceutical industry due to availability trained and skilled manpower, good infrastructure as well as research and development facilities. Hyderabad accounts for around one-third of Indias total bulk production, is considered as the drug capital of the country. It provides plenty of opportunities for entrepreneurs as govt, taken a decision for the establishment of pharma city near Visakhapatnam.

4) Mines and Minerals:
Andhra Pradesh is the second largest store house of minerals resources in India. It includes vast deposits of coal, limestone, slabs, oil and natural gas,, manganese, iron ore etc. A wide range of these minerals use in fertilisers, ceramics, abrasives, glass, foundry, oil well drilling filters and pigments. Besides the state has immense potential of untapped and under tapped minerals and provides opportunities for entrepreneurs.

5) Agriculture and Forestry :
Agriculture is the main occupation for the people in A.R Rice is the major crop in the state. Other important crops are lower bajra, pulses, cluster etc. Entrepreneurs may have many opportunities to establish units based on these products. Entrepreneurs make use of recently announced industrial policy of A.R and explore the possibilities of establishing agro based units.

6) Tourism:
A.P is truly a land of beauty and it represents Indian culture and heritage. Beaches, wild life and forests, forts, historical monuments, national parks, holy temples with architectural beauty. It provide many opportunities to entrepreneurs to start ventures related to tourism services.

7) Fisheries:
The state has stretched over the coastal area of Bay of Bengal. A vast coastal area is the main reason for sea food. The aqua culture around the coastal proved to be a rich source for sea food, which occupies a major share in exports from the state. The entrepreneurs may examine the various opportunities and explore the possibilities of setting units based on the sea foods and export the same.

Short Answer Questions

Question 1.
Explain the meaning of entrepreneur.
Answer:
The word entrepreneur is derived from the French verb ‘entrepredre’, which means to undertake. An entrepreneur may be referred to such a single person or group of persons who promote a new enterprise by collecting various factors of production and bearing the risks arising out of such venture.

Question 2.
Write one definition of entrepreneur.
Answer:
According to Peter F.Drucker an entrepreneur is one who “searches for change, responds to it and exploits opportunities and the innovation is the specified tool of an entrepreneur”.

Question 3.
Write one definition of entrepreneurship.
Answer:
In a conference on entrepreneurship held in United States, the term entrepreneurship was defined as “Entrepreneurship is an attempt to create value through recognition of business opportunity, the management of risk-taking appropriate to the opportunity and through the-communication and management skills to mobilise human, financial and material resources necessary to bring a project to function”.

Question 4.
Explain the any one characteristic of an entrepreneur. [A.P. Mar. 17]
Answer:
The following are the some of the important characteristics that every successful entrepreneur must possess.

1) Innovation:
Innovation is an important characteristic of an entrepreneur in modem business. The entrepreneur makes arrangements for introducing innovations which help in increasing production on the one hand and reducing costs on the other.

2) Risk-taking :
Risk taking is another feature of an entrepreneur. He has to pay to all the other factors of production in advance. There are chances that he may be rewarded with a handsome profit or he may suffer a heavy loss. Therefore, the risk bearing is the final responsibility of an entrepreneur.

Question 5.
Explain the any one function of an entrepreneur.
Answer:
An entrepreneur performs all the functions necessary right from the generation of an idea and upto setting up an enterprise. The following are the functions of entrepreneur.

Decision making :
An entrepreneur as a decision maker takes various decisions regarding the following :

1. Ascertaining the objective of the enterprise
2. Sources of finance
3. Product mix
4. Pricing policies
5. Promotion strategies
6. Appropriate technology or new equipments etc.

Question 6.
Write the types of entrepreneurs. [A.P. Mar. 17]
Answer:
Entrepreneurs are classified into four types. They are :

1. Innovating entrepreneurs.
2. Initiative entrepreneurs.
3. Fabian entrepreneurs.
4. Drone entrepreneurs.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry

Very Short Answer Questions

Question 1.
What is an interface? Give one example.
Answer:
The interface is normally a few molecules thick but its area depends on the size of the particles of bulk.
Interface is represented by separating the bulk phases by hyphen (-) (or) a slash (/).
Eg : Interface between a solid and a gas may be represented by solid-gas (or) solid/gas.

Question 2.
What is adsorption ? Give one example.
Answer:
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Question 3.
What is absorption ? Give one example.
Answer:
Absorption : The uniform distribution of a substance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Question 4.
Distinguish between adsorption and absorption. Give one example of each.
Answer:
Adsorption
1. The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg :Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Absorption
1) The uniform distribution of a susbstance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Question 5.
The moist air becomes dry in the presence of silica gel. Give reason for this.
Answer: The moist air becomes dry in presence of silica gel because the water molecules present in air get adsorbed on the surface of the gel.

Question 6.
Methylene blue solution when shakes with animal charcoal gives a colourless filtrate on filteration. Give the reason.
Answer:
Methylene blue solution (organic dye) when shaken with animal charcoal gives a colourless filtrate on filtration. The molecules of the dye from the solution are adsorbed on the charcoal.

Question 7.
A small amount of silica gel and a small amount of anhydrous calcium chloride are placed separately in two corners of a vessel containing water vapour. What phenomena will occur ?
Answer:
A small amount of silica gel and a small amount of anhydrous CaCl₂ are placed separately in two comers of a vessel of water vapour. Water vapours are is absorbed by anhydrous CaCl₂ but adsorbed by silical gel.

Question 8.
What is desorption ?
Answer:
Desorption: The process of removing aa adsorbed substance from a surface on which it is adsorbed is called desorption.

Question 9.
What is sorption ?
Answer:
In case of some substances both adsorption and absorption takes place. This phenomenon is called sorption.

Question 10.
Amongst adsorption, absorption which is a surface phenomena and why ?
Answer:

1. Adsorption is a surface phenomenon.
2. In adsorption the concentration of the adsorbate increases only at the surface of the adsorbent. While in absorption the concentration is uniformly distributed throughout the bulk of the compound.

Question 11.
What is the name given to the phenomenon when both absorption and adsorption take place together ?
Answer:
The name of the phenomenon when both adsorption and absorption take place together is sorption.

Question 12.
Chalk stick dipped in an ink solution exhibits the following :
a) The surface of the stick retains the colour of the ink.
b) Breaking the chalk stick,-it is found still white from inside.Explain the above observations.
Answer:
a) Chalk stick dipped in an ink solution, the surface of the stick retains the colour of the ink due to adsorption of coloured molecules in the ink.
b) Chalk stick dipped in an ink solution, the surface of the stick retains the colour of the ink due to adsorption of coloured molecules in the ink while the solvent of the in k goes deep into the stick due to absorption so on breaking the chalk stick, it is found to be white inside.

Question 13.
What are the factors which influence the adsorption of a gas on a solid ?
Answer:
Adsorption of a gas on solids is influenced by the following factors.
a) Surface area of the adsorbent, b) Nature of the gas, c) Pressure of the adsorbate, d) Temperature.

Question 14.
Why is adsorption always exothermic ?
Answer:
During the adsorption, there is always a decrease in residual forces of the surface. There is decrease in surface energy which appears as heat.
∴ Adsorption is always Exothermic.

Question 15.
Give the signs of ∆H & ∆S, when ammonia gas gets adsorbed on charcoal.
Answer:
When NH₃ gas gets adsorbed on charcoal the sign of ∆H is negative and the sign of ∆S is also negative.

Question 16.
How many types of adsorption are known ? What are they ?
Answer:
Adsorption process is divided into two types.
1) Physisorption 2) Chemisorption

Question 17.
What types of forces are involved in physisorption of a gas on solid ?
Answer:
Weak Vander waals forces are involved in physisorption of a gas on solid.

Question 18.
What type of interaction occurring between gas molecules and a solid surface is responsible for chemisorption of the gas on solid.
Answer:
Chemical bonds occur between gas molecules and a solid surface, is responsible for chemisorption of the gas on solid.

Question 19.
Why chemisorption is called activated adsorption ?
Answer:
Chemisorption involves a high energy of activation. Hence it is referred as activated adsorption.

Question 20.
What is the difference between physisorption and chemisorption ?
Answer:
1) In physisorption the forces present between adsorbate and adsorbent are weak vander . waal’s forces.
2) In chemisorption the forces present between adsorbate and adsobent are chemical bonds (valency forces).

Question 21.
Out of physisorption and chemisorption, which can be reversed ?
Answer:
Out of physisorption and chemisorption, phyisorption is reversed (reversible).

Question 22.
How is adsorption of a gas related to its critical temperature ?
Answer:
Higher the critical temperature, greater is the case of liquefaction of a gas. Then the extent of adsorption will be high.

Question 23.
The critical temperature of SO₂ is 630 K and that of CH₄ is 190 K. Which is adsorbed easily on activated charcoal ? Why ?
Answer:
Given the critical temperature of SO₂ is 630 K and that of CH₄ is 190 K.
SO₂ gas is adsorbed easily on activated charcoal because of higher critical temperature. Higher the critical temperature of gas, extent of adsorption is high.

Question 24.
Easily liquefiable gases aye readily adsorbed on solids. Why ?
Answer:
Easily liquefiable gases are readily adsorbed on solids because vander waal’s forces are stronger near the critical temperatures.

Question 25.
Amongst SO₂, H₂ which will be adsorbed more readily on the surface of charcoal and why ?
Answer:
Among SO₂, H₂ gases, SO₂ gas adsorb more readily on the surface of charcoal because SO₂ has high critical temperature (630 K) than dihydrogen (33K).

Question 26.
Compare the enthalpy of adsorption for physisorption and chemisorption.
Answer:

1. In physical adsorption enthalpy of adsorption is low (20 – 40 KJ/mole)
2. In case of chemical adsorption enthalpy of adsorption is high (80 – 240 KJ/mole)

Question 27.
What is the magnitude of enthalpy of physical adsorption ? Give reason for this magnitude.
Answer:
The magnitude of enthalpy of physical adsorption is low 20 – 40 KJ ,/mole. This is due to the presence of weak vander waal’s forces between gas molecules and solid surface.

Question 28.
What is the magnitude of enthalpy of chemisorption ? Give reason for this magnitude.
Answer:
The magnitude of enthalpy of chemical adsorption is high 80 – 240 KJ /mole. This is due to the presence of chemical bonds between gas molecules and solid surface.

Question 29.
Give any two applications of adsorption.
Answer:
Applications of adsorption :
a) Separation of inert gases : Different noble gases adsorb at different temperatures on coconut charcoal. By this principle (adsorption) mixture of noble gas is separated by adsorption on coconut charcoal.
b) Gas masks : Gas mask is a device which consists of activated charcoal (or) mixture of adsorbents is used by coal miners to adsorb poisonous gases during breathing.

Question 30.
Why physisorption suffers from lack of specificity ?
Answer:
Physisorption suffers from lack of specificity.
Explanation: A given surface of an adsorbent doesnot show any preference of a particular gas as the vander waal’s forces are universal.

Question 31.
What is an adsorption isotherm ? Write the equation of Freundlich adsorption isotherm.
Answer:
Adsorption Isotherm : The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve known as adsorption isotherm.
Freundlich adsorption isotherm equation is \(\frac{x}{m}\) = k. P^1/n
x = mass of the gas adsorbed
m = mass of the adsorbent
P, k and n are constants.

Question 32.
In the Freundlich adsorption isotherm, mention the conditions under which, following graph will be true ?

Answer:
In the above graph extent of adsorption does not depends on the pressure.
When \(\frac{1}{n}\) = 0, \(\frac{x}{m}\) = constant, the adsorption is independent of pressure.

Question 33.
What role does adsorption play in heterogenous catalysis ?
Answer:
In heterogeneous catalysis, adsorption of reactants on the solid surface of the catalysts increases the rate of reaction.
Eg : Manufacturing of NH₃ using Fe-as catalyst [Haber’s process].

Question 34.
What is the role of MnO₂ in the preparation of O₂ from KClO₃ ?
Ans. The chemical equation for the preparation of O₂ from KClO₃ is

MnO₂ increases the rate of reaction. Hence it is a catalyst for the above reaction.

Question 35.
Define “promoters” and “poisons” in the phenamenon of catalysis ?
Answer:
Promoters : The substances with enhance the activity of catalyst are known as promoters. Poisons : The substances which decrease the activity of a catalyst are known as poisons.

Question 36.
What is homogeneous catalysis ? How is it different from heterogeneous catalysis ?
Answer:
Homogeneous Catalysis : The catalysis in which reactants and catalyst are in same phase is called Homogeneous catalysis.
In case of heterogeneous catalysis, catalyst and reactants are present in different phases where as in case of homogeneous catalysis catalyst and reactants are present in same phase.

Question 37.
Give two examples for homogeneous catalytic reactions.
Answer:
The following are the examples for homogeneous catalytic reactions.

Question 38.
Give two examples for heterogeneous catalysis.
Answer:
The following are the examples of heterogeneous catalytic reactions.

Question 39.
Give two examples which indicate the selectivity of heterogeneous catalysis.
Answer:
The following reactions indicate the selectivity of heterogeneous catalysis.
Starting with H₂ and CO, and using different catalysts, we get different products.

The action of a catalyst is highly selective in nature. A substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 40.
Why zeolites are treated as shape selective catalysts ?
Answer:

• The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysts.
• Zeolites are shope selective catalysts because of their honey comb-like structure. They are microporous alumino silicates with 3-dimensional net work of silicates.

Question 41.
Which zeolite catalyst is used to convert alcohols directly into gasoline ?
Answer:
Zeolite ZSM – 5 is used to convert alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Question 42.
What are enzymes ? What is their role in human body ?
Answer:
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.
These act as specific catalysts in biological reactions.
These catalyse the numerous reactions that occur in the bodies of animals and plants to maintain the life process.

Question 43.
Can catalyst increase the yield of reaction ?
Answer:
Catalyst does not increase the yield of reaction. Catalyst just speed up the product formation.

Question 44.
Name any two enzyme catalyzed reactions. Give the reactions.
Answer:
1) Inversion of Cane Sugar :

2) Decomposition of urea into ammonia and CO₂:

Question 45.
Name the enzymes obtained from soyabean source.
Answer:
The enzyme obtained from soyabean source is urease.

Question 46.
Name the enzymes used in
a) Decomposition of urea into ammonia.
b) Conversion of proteins into peptides in stomach.
Answer:
a) The enzyme used in the decomposition of urea into ammonia is urease.
b) The enzyme used in conversion of proteins into peptides in stomach is pepsin.

Question 47.
What enzymes are obtained from yeast ?
Answer:
The enzymes obtained from yeast are invertase, zymase and maltase.

Question 48.
At what ranges of temperature and pH, enzymes are active ?
Answer:
The optimum temperature range for enzymatic activity is 298 – 310 K. Human body temperature being 310 K is suited for enzyme catalysed reactions. The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH, which lies between 5-7.

Question 49.
Represent diagrammatically the mechanism of the enzyme catalyis.
Answer:

Question 50.
Name any two industrially important heterogeneous catalytic reactions mentioning the catalysts used.
Answer:
i) Manufacturing of NH₃ by Haber’s process

ii) Hydrogenation of vegetable oils in presence of finely divided nickel as catalyst.

The above mentioned are industrially important heterogeneous catalytic reactions.

Question 51.
What is a colloidal solution ? How is it different from a true solution with respect to dispersed particle size and homogeneity ?
Answer:
A heterogeneous system in which one substance is dispersed as large particles in another substance is called colloidal solution.

• In colloidal solution the particle size of the dispersed phase is of the order of 1 mμ – 1 μ where as in true solution the particle size of the solute are the order of mp or less.
• Colloidal solution is a heterogeneous binary system where as true solution is a homogeneous binary system.

Question 52.
Name the dispersed phase and a dispersion medium in the following colloidal systems

1. fog
2. smoke
3. milk.

Answer:

1. Fog : Dispersed phase → liquid
   Dispersion medium → gas
2. Smoke : Dispersed phase → carbon particles (solid)
   Dispersion medium → Air
3. Milk : Dispersed phase → liquid fat
   Dispersion medium → water

Question 53.
What are lyophilic and lyophobic sols ? Give one example for each type.
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.

The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 54.
Explain the terms with suitable examples .

1. aerosol
2. hydrosol.

Answer:

1. Aerosol: The colloidal solution in which dispersed phase is solid and dispersion medium is gas is called an aerosol.
   Eg : Smoke, dust.
2. Hydrosol: In a colloidal solution dispersion medium is water them it is called aquasol (or) hydrosol.
   Eg : Milk, Gold sol.

Question 55.
Explain why lyophilic colloids are relatively more stable than lyophobic colloids ?
Answer:

• Lyophilic colloids are reversible and are quite stable. These are not coagulated.
• Lyophobic sols are irreversible and need stabilizing agents for their stabilization. These sols are readily precipitated on the addition of small amounts of electrolytes.

Question 56.
Give two examples of colloidal solutions of liquids dispersed in solid. What is the name given to the colloidal solution ?
Answer:
Cheese, butter, jellies are examples of colloidal solutions of liquids dispersed in solid. The name of the colloidal solution given to this type is Gels.

Question 57.
What is the difference between multimolecular and macromolecular colloids ? Give one example for each.
Answer:

• In multimolecular colloids, a large number of atoms (or) small molecules of the dispersed phase aggregate together to form species in the colloidal range.
• Macro molecules in suitable solvents form solutions in which the sizes of the macromolecules are in the colloidal range.

Question 58.
What are micelles ? Give one example.
Answer:
Micelles : Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.
Eg : Stearate ions (C₁₇H₃₅COO^–) associate together in high concentration, in a solution of soap in water and forms a micelle.

Question 59.
How do micelles differ from a normal colloidal solutions ?
Answer:
Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.

• hese are also known as associated colloids.
• These colloids have both lyophilic and lyophobic parts.
• Micelles may contain as many as 100 or more of normal molecules.
• On dilution these colloids revert back to individual electrolytes.

Question 60.
Give two examples of associated colloids.
Answer:
Surface active agents such as soaps and synthetic detergents are examples of associated colloids.

Question 61.
Can the same substance act both as colloid and crystalloid ?
Answer:
Yes, Micelles, (associated colloids) act as normal strong electrolytes at low concentrations and exhibit colloid behavior at high concentrations.

Question 62.
Give two examples of lyophobic sols.
Answer:
Metal sols and metal suiphide sols are examples of lyophobic colloids

• Gold sol is lyophobic colloid

Question 63.
Give examples of’colloidal system of

1. Liquid In solid
2. Gas in solid.

Answer:

1. Cheese, butter, jellies are examples of liquid in solid type colloidal system.
2. Foam rubber, pumice ‘stone are examples of gas in solid type of colloidal system.

Question 64.
What type of substances form lyophobic sols?
Answer:
Substances like metals, their sulphides.donot form colloidal sol simply by mixing with the dispersion medium. These form colloids by special methods, These sols are lyophobic colloids.

Question 65.
What is critical micelle concentration (CMC) and kraft temperature (T_k) ?
Answer:
The formation of micelles takes place only above a particular temperature called Kraft temperature (T_k) and above a particular concentration called Critical micelle concentration (CMC).

Question 66.
Why lyophobic colloids are called irreversible colloids ?
Answer:

• Lyophobic colloids are coagulated by the addition of small amounts of electrolytes.
• The precipitates does not give back the colloidal sol by same addition of the dispersion medium.
• Hence lyophobic sols are called irreversible sols.

Question 67.
How a colloidal sol of arsenous sulphide is prepared ?
Answer:
Colloidal sol of arsenous sulphide is prepared by the double decompostion of As₂O₃ and H₂S as follows.

Question 68.
What is Peptization ?
Answer:
Peptization : The process of converting a precipitate into colloidal sol by shaking it with the dispersion medium in the presence of a small amount of electrolyte is called Peptization.

Question 69.
What is dialysis ? How is dialysis can be made fast ?
Answer:
Dialysis : The process of removing a dissolved substance from a colloidal solution using a suitable membrane is called dialysis.
Dialysis is made faster by applying an Emf. This is known as Electrodialysis.

Question 70.
What is collodion solution ?
Answer:
Collodion solution : Collodion is a 4% solution of nitro cellulose in a mixture of alcohol and ether.

Question 71.
How an ultrafilter paper is prepared from ordinary filter paper ?
Answer:
Ultra filter paper is prepared by soaking the filter paper in collodion solution, hardening by formaldehyde and then finally dried.

Question 72.
What is Tyndall effect ?
Answer:
Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light. A true solution does not show Tyndall effect.

Question 73.
What conditions is tyndall effect observed ?
Answer:
Tyndall effect is observed only when

1. The diameter of the dispersed particles is not much smaller than the wave length of the light used.
2. The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Question 74.
Can Tyndall effect be used to distinguish between a colloidal solution and a true solution ? Explain.
Answer:
Tyndall effect is used to distinguish between a colloidal solution and a true solution.

• True solution placed in dark and is observed in the direction of the light passed beam passing through it. It appears clear and it is observed from a direction at right angles to the direction of the light beam, it appears perfectly dark.
• Colloidal solution viewed in the same way may appear reasonably clear in the direction of light but they show a mild to strong opalescence when viewed at right angles to the direction of light.

Question 75.
Sky appears blue in colour. Explain.
Answer:
Dust particles along with water vapour suspended in air scatter blue light which reaches our eyes and hence the sky looks blue to us.

Question 76.
What is Brownian movement.
Answer:
Brownian movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig-zag motion.
This rapid motion of colloidal particles is called Brownian movement.

This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

Question 77.
What is the main cause for charge on a colloidal solution ?
Answer:
The charge on the colloidal particles is due to electron capture by sol particles during electro dispersion of metals , and due to preferential adsorption ions from solution (or) due to formulation of electrical double layer.

Question 78.
What is electrokinetic potential or zeta potential ?
Answer:
In a colloidal sol the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.

Question 79.
Write the chemical formula of positively charged and negatively charged hydrated ferric oxide colloidal solutions.
Answer:

• Formula of positively charged hydrated ferric oxide is Fe₂O₃.xH₂O/Fe⁺³
• Formula of negatively charged hydrated ferric oxide is Fe₂O₃.xH₂0/OH^–

Question 80.
Give the order of coagulating power of Cl, SO₄^2-, PO₄^3- in the coagulation of positive sols.
Answer:
The order of coagulating ability’of given ions with positive sols is PO₄^3- > SO₄^2- > Cl^–

Question 81.
Amongst Na⁺, Ba²⁺, Al³⁺, which coagulates negative sol readily and why ?
Answer:
The order of coagulating ability of given ions with negative sols is Al⁺³ > Ba⁺² > Na⁺

Question 82.
A colloidal solution of AgI is positively charged when prepared from a solution containing excess of Ag⁺ ions and negatively charged when prepared from a solution containing excess of I^– ions Explain.
Answer:
When a dilute AgNO₃ solution is added to a dilute KI solution taken in excess, the precipitated Agl adsorbs I^– ions present in excess and a negatively charged AgI colloidal solution is formed. However when dilute KI solution is added to dilute AgNO₃ solution taken in excess, AgI precipitate adsorbs Ag⁺ ions present in excess and a positively charged AgI colloidal solution is formed. Generally the ion common to dispersed particle is adsorbed. .
AgI/I^–
Negatively charged
AgI/Ag⁺
Positively charged.

Question 83.
What is electrophoresis ?
Answer:
When electric potential is applied across two platinum electrodes dipping in a colloidal solution,-the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Question 84.
What is electro osmosis ?
Answer:
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in opposite direction. This phenomenon is termed “electro osmosis”.

Question 85.
What is coagulation ?
Answer:
The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If this charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.

Question 86.
Define flocculation value.
Answer:
The minimum concentration of an electrolyte in millimoles per litre required to cause coagulation of a sol in two hours is called “coagulating value” (or) flocculation value.

Question 87.
State Hardy – Schulze rule.
Answer:
Greater the valence of the coagulating ion added, the greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Questions 88.
Coagulation takes place when sodium chloride solution is added to a colloidal solution of hydrated ferric oxide. Explain.
Answer:
When NaCl solution is added to the colloidal solution of hydrated ferric oxide coagulation takes place (precipitation formed). .
The reason is that colloidal particles interact with ions (Na⁺, Cl^–) carrying charge opposite to that present on themselves. This causes neutralisation of charges leading to their coagulation.

Question 89.
How are lyophobic solutions protected from phenomenon of coagulation.
Answer:
Lyophobic sols are protected from the phenomenon of coagulation by adding suitable lyophilic sol.

Question 90.
What is protective colloid ?
Answer:
Lyophilic colloids used for the prevention of coagulation of lyophobic colloids are called protective colloids.

• Lyophilic colloids protect the lyophobic colloids.

Question 91.
What is an emulsion ? Give two examples. [A.P. Mar. 17]
Answer:
Emulsion: The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called emulsion.
Eg : Milk, Vanishing cream, Cold cream.

Question 92.
How emulsions are classified ? Give one example for each type of emulsion. [A.P. Mar. 17]
Answer:
Emulsion : A dispersion of finely divided droplets of a liquid in another liquid medium is called’emulsion’.
Ex: Milk.
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.
Classification of emulsions : Emulsions are “classified into two classes. These are
a) oil in water (O/W) and
b) water in oil (W/O), (O = Oil; W = Water).

These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.
a) Oil in Water (O/W) type emulsions : In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex : Milk, liquid, fat (oil) in water.
Vanishing cream; fat in water.

b) Water in Oil (W/O) type emulsions : In this type of emulsions, the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil
Cold cream : water in fat.

Question 93.
What is an emulsifying agent ?
Answer:
Emulsifying agent: The third substance which is added in small amounts to an emulsion to stabilize the emulsion is called emulsifying agent.

Question 94.
What is demulsification ? Name two demulsifiers.
Answer:
Demulsification : The separation of an emulsion into constituent liquids is known as demulsification.

Question 95.
How is artificial rain produced ?
Answer:
Artificial rain is produced by throwing electrified sand (or) spraying a sol carrying charge opposite to the one on clouds from an aeroplane.

Question 96.
Bleeding from fresh cut can be Stopped by applying alum. Give reasons.
Answer:
Bleeding from fresh cut can be stopped by applying alum. This is due to styptic action of alum which coagulates the blood and stops further bleeding.

Question 97.
Deltas are formed at the points where river enters the sea. Why ?
Answer:
Deltas are formed at the points where river enters the sea.
Explantation : River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay resulting in its deposition of clay with the formation of delta.

Question 98.
Name any two applications of colloidal solutions.
Answer:
Applications of colloidal solutions :
Rubber: Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation.
Industrial Products : Paints, inks, synthetic plastics, rubber, graphite, lubricants, cement, etc., are all colloidal in nature.

Question 99.
How can aerial pollution by colloidal particles of smoke be prevented ? Explain.
Answer:
Aerial pollution by colloidal particles of smoke be prevented by electrical precipitation of smoke. Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust, etc., in air. The smoke, before it comes out from the chimney, is led through a precipitator containing plates having a charge opposite to that carried by smoke particles. The particles on coming in contact with these plates lose their charge and get precipitated. The particles thus settle down on the floor of the chamber. The precipitator is called cottrell precipitator.

Question 100.
Alum is used to purify water obtained from natural sources. Explain.
Answer:
The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

Question 101.
Why medicines are more effective in colloidal state ?
Answer:
Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Question 102.
How rubber is obtained from latex ?
Answer:
Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation.

Question 103.
Name the type of emulsion to which milk belongs.
Answer:
Milk is a oil in water type of emulsion.
Dispersed phase : liquid fat.
Dispersion medium : water.

Short Answer Questions

Question 1.
What is adsorption ? Discuss the mechanism of adsorption of gases on solids.
Answer:
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.
Mechanism of Adsorption :

• Adsorption arises due to the fact that the surface particles of the adsorbent are not in the same environment as the particles inside the bulk.
• Adsorbent possess unbalanced or residual attractive forces. These forces are responsible for attracting the adsorbate molecules on adsorbent surface.
• The extent of adsorption increases with increase in surface area of adsorbent at a given temperature and pressure.
• During the adsorption, there is always a decrease in residual forces of the surface. There is decrease in surface energy which appears as heat. So, Adsorption is always Exothermic.
• Adsorption is accompanied by decrease in enthalpy as well as decrease in entropy of system. .
• As the adsorption proceeds AH becomes less and less negative, ultimately, AH becomes equal to TAS and AG becomes zero. At this state equilibrium is attained.

Question 2.
What are different types of adsorption ? Give any four differences between characteristics of these different types. [T.S. Mar. 19, 15; A.P. Mar. 15]
Answer:
Adsorption process is divided into two types.
1) Physisorption
2) Chemisorption.
Distinguishing characteristics of Physisorption and Chemisorption are given in the following table:
Physisorption

1. This process is weak, due to vander Waal’s forces.
2. The process is reversible.
3. This is a quick process, i.e., takes place quickly.
4. The process decreases with increase of temperature.
5. This is a multilayered process.
6. The process depends mainly on the nature of the adsorbent.

Chemisorption

1. This process is strong, due to chemical forces.
2. The process is irreversible.
3. This is a slow process.
4. The process increases with increase of temperature.
5. This is a unilayered process.
6. The process depends both on the nature of adsorbent and adsorbate.

Question 3.
What do you understand by the terms given below (a) absorption (b) Adsorption (c) Adsorbent and Adsorbate
Answer:
Absorption : The uniform distribution of a susbstance through out the bulk of the solid .substance is known as absorption. Eg: Chalk stick dipped in ink.
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg: Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Adsorbent: The substance on whose surface the adsorption occurs is known as adsorbent.

Adsorbate : The substance whose molecules get adsorbed on the surface of the adsorbent is known as adsorbate.
Eg: In the adsorption of acetic acid by charcoal, acetic acid is adsorbate and the charcoal is adsorbent.

Question 4.
Adsorption of a gas on the surface of solid is generally accompanied by decrease in entropy. Still it is a spontaneous process. Explain.
Answer:
Adsorption is accompanied by decrease in enthalpy as well as decrease in entropy of the system.

• For a process to be spontaneus there is a decrease in Gibbs Energy. .
• On the basis of equation ∆G = ∆H – T∆S; ∆G = -Ve
•  If AH has sufficiently high negative value and T∆S is positive.
• In an adsorption process which ¡s spontaneous ∆G becomes negative by combining above.
• As the adsorption proceeds ∆H becomes less and less negative ultimately, ∆H becomes equal to T∆S and ∆G becomes zero. At this state equilibrium is attained.

Question 5.
How can the constants k and n of the Freundlich adsorption equation be calculated?
Answer:
Freundlich adsorption isotherm equation is
\(\frac{x}{m}\) = k. P =, x/m = Extent of adsorption ⇒ P = Pressure
k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.

Applying logarithm to the above equation
log \(\frac{x}{m}\) = log k + \(\frac{1}{n}\) log P.

• A graph is plotted taking log \(\frac{x}{m}\) on y – axis and log P on x – axis. 1f the graph is a straight line then Freundlich isotherm is valid.
• The slope of the straight line gives \(\frac{1}{n}\) value.
• The intercept on the y-axis gives value of log k.
• \(\frac{1}{n}\) has values between 0 and 1
• When \(\frac{1}{n}\), = 0, \(\frac{x}{m}\) = constant, the adsorption is independent of pressure
  \(\frac{1}{n}\) = 1, \(\frac{x}{m}\) = kP i.e., \(\frac{x}{m}\) ∝ p.

Question 6.
How does the extent of adsorption depend upon
a) Increasing the surface area per unit mass of adsorbent.
b) Increasing temperature of the system.
c) Increasing pressure of the gas.
Answer:
a) The extent of adsorption increases by increasing the surface area per unit mass of adsorbent.
b) The extent of adsorption decrease with an increase in temperature.
c) The extent of adsorption increase with increasing pressure of the gas.

Question 7.
What is catalysis ? How is catalysis classified ? Give two examples for each type of catalysis. [A.P. Mar. 16] [Mar. 14]
Answer:
Catalysis : A substance which alters the rate of a chemical reaction without itself being consumed in the process, is called a catalyst.
The action of catalyst in altering the rate of a chemical reaction is called catalysis.

Types of catalysis: Catalysis is classified into two types as
a) Homogeneous catalysis and b) Heterogeneous catalysis.

a) Homogeneous catalysis : The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

b) Heterogeneous catalysis : The catalytic process in which the catalyst is present in a phase different from that of the reactants is known as heterogeneous catalysis.

Question 8.
Discuss the mechanism involved in adsorption of heterogeneous catalysis.
Answer:
Adsorption theory of heterogeneous catalysis :
Adsorption theory explains the mechanism of heterogeneous catalysis.

• The modern adsorption theory is the combination of the intermediate compound formation theory and the old adsorption theory.
•  The mechanism of catalysis involves five steps.
• Diffusion of the reactants to the surface of the catalyst.
• Adsorption of the reactant molecules on the surface of the catalyst.
• Occurrence of chemical reaction on the catalyst’s surface through formation of an . intermediate.
• Desorption of reaction products from the catalyst surface, and thereby making the surface available again for more reaction to occur.
• Diffusion of reaction products away from the catalyst’s surface.

Question 9.
Discuss some features of catalysis by zeolites.
Answer:
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysts.

• Zeolites are shope selective catalysts because of their honey comb-like structure. They are microporous alumino silicates with 3-dimensional net work of silicates.
• Zeolites are microporous alumino silicates with dimensional network of silicates in which some Si-atoms are replaced by aluminium atoms giving Al-o-Si frame work.
• The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as on the pores and cavities of the zeolites.
• Zeolites are found in nature and also synthesized for catalytic selectivity.

Uses:

• Zeolites are used as catalysts in petrochemical industries for cracking.
• Zeolite ZSM – 5 is used to convert alcohol directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Question 10.
Give brief account of mechanism of enzyme catalysis with suitable diagrams.
Answer:
Mechanism of enzyme catalysis :

The enzyme – catalysed reactions may be considered to proceed in two steps.
Step 1 : Binding of substrate to enzyme to form an activated complex (ES^≠).
E + S → E S^≠
Step 2 : Decomposition of the activated complex to form product.
E S^≠ → E + P

Question 11.
Discuss the factors that influence the catalytic activity of enzymes.
Answer:
Factors influencing the catalytic activity of enzymes :

• Highly specific nature: One catalyst catalyses only one reaction. Each enzyme is specific ‘ for a given reaction.
  Eg : The enzyme urease catalyses the hydrolysis of urea only and not any other amide.
• Highly active under optimum temperature: The rate of an enzyme reaction is maximum at a temperature called optimum temperature.
• The optimum temperature range for enzymatic activity is 298 – 310 K.
• Highly active under optimum pH : The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH which lies between 5 – 7.
• Increasing activity in presence of activators and co-enzymes : The enzymatic activity is increased in presence of co-enzymes (or) in presence of activators such as Na⁺, Mn⁺² etc.,
•  Influence of inhibitors and poisons: The inhibitors (or) poisons interact with the active functional groups on the enzyme surface and reduce or completely destroy the catalytic activity.

Question 12.
Name any six enzyme catalysed reactions. [A.P. Mar. 19]
Answer:
i) Inversion of Cane sugar : Enzyme : Invertase

Question 13.
What do you mean by activity and selectivity of catalysts ?
Answer:
Activity :
The ability of a catalyst in increasing the rate of reaction is defined as activity of catalyst.

• The activity of a catalyst depends upon the strength of chemisorption to a large extent.
• The reactants must get adsorbed reasonably strongly on to the catalyst to become reactive.
  Eg : The catalystic activity increases from Group – 5 to Group – 11 for hydrogenation reactions.
  The maximum activity being shown by 7 – 9 group metals.
  

Selectivity:
The selectivity of a catalyst is its-ability to direct a reaction to form specific products. The following reactions indicate the selectivity of heterogeneous catalysis.
Starting with H₂ and CO, and using different catalysts, we get different products.

The action of a catalyst is highly selective in nature. A substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 14.
How are colloids classified on the basis of physical states of components ?
Answer:
On the basis of physical states of dispersed phase and dispersion medium colloidal solutions are classified into eight types.

Question 15.
How are colloids classified on the basis of nature of the dispersion medium ?
Answer:
On the basis of nature of dispersion medium colloids are classified as follows.

•  If the dispersion medium is ‘air’, the sols are called aerosols.
  Eg : Smoke.
• If the dispersion medium is ‘water’ the sols are called Hydrosol (or) aquasol.
  Eg : starch solution.
• If the dispersion medium is ‘alcohol’ then the sols are called alcosols.

Question 16.
How are colloids classified on the basis of interaction between dispersed phase and dispersion medium ?
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex: Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 17.
What is the difference between a colloidal sol, gel, emulsion and a foam?
Answer:

Question 18.
What are lyophilic and lyophobic sols? Compare the two terms in terms of stability and reversibility.
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 19.
Name a substance whose molecules consist of lyophilic as well as lyophobic parts. Give its use in our daily life.
Answer:

• Associated colloids (or) micelles contains both lyophobic and lyophillic parts.
• Examples are soaps and synthetic detergents.
• The cleaning action of soap is due to the fact that soap molecule form a micelle around the oil droplet.

Question 20.
Describe Bredig’s arc method of preparation of colloids with a neat diagram.
Answer:
Electrical disintegration or Bredig’s Arc method : This process involves dispersion as well as condensation. Colloidal sols of metals such as gold, silver, platinum, etc., are prepared by this method. In this method, electric arc is struck between electrodes of the metal immersed in the dispersion medium. The intense heat produced vapourises the metal, which then condenses to form particles of colloidal size.

Question 21.
Name any four examples of preparation of colloids by chemical methods with necessary chemical equations.
Answer:
Chemical methods : Colloidal solutions are prepared by chemical reactions leading to the formation of species by double decompostion, oxidation, reduction or hydrolysis. These species then aggregate leading to the formation of sols.

Question 22.
Describe the purification of colloidal solutions by the phenomenon of dialysis with a neat diagram. [A.P. Mar. 18]
Answer:
Colloidal solutions generally contain excess amount of electrolytes and some other soluble impurities. It is necessary to reduce the concentration of soluble impurities to a requisite minimum. The impurities presence required in traces.
“The process used for reducing the amount of impurities to a requisite minimum is known as purification of colloidal solution”.

Purification of colloidal solution by Dialysis :
Dialysis : The process of removing a dissolved substance from a colloidal solution using a suitable membrane is called dialysis.

• In a true solution particles can pass through animal membrane (or) cellophane sheet (or) parchment paper but not colloidal particles.
• The apparatus used for the dialysis is called dialyser.
• A bag of suitable membrane containing the colloidal solution is suspended in a vessel containing a continuously flowing water.
• The molecules and ions diffuse through the membrane into the water and pure colloidal solution is left behind in the bag.

Question 23.
Explain the formation of micelles with a neat sketch.
Answer:
Mechanism of micelle formation: Let us take the example of soap solution. Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOC^– Na⁺ (e.g., sodium strearate CH₃ (CH₂)₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na⁺ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ‘tail’) which is hydrophobic (water repelling), and a polar group COO^–(also called polar-ionic ‘head’), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. But at critical micelle concentration, the COO^– irons are pulled into the bulk of the solution and are aggregated to form a spherical shape with their hydrocarbon chains pointing towards the centre of the . sphere with COO^– part remaining outward on the surface of the sphere. The aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 simple molecules.

Similarly, in case of detergents, e.g., sodium laurylsulphate. CH₃ (CH₂)₁₁ SO₃^–Na⁺, the polar group is -SO₃^– along with the long hydrocarbon chain. Hence, the mechanism of micelle formation here also is same as that of soaps.

Question 24.
Action of soap is due to emulsification and micelle formation. Comment.
Answer:
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.

Cleaning action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

Soap thus, functions as an emulsifying agent for the water dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 25.
Explain the phenomenon of Brownian movement giving reasons for the occurrence of this phenomena.
Answer:
Brownian movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig – zag motion.
This rapid motion of colloidal particles is called Brownian movement. This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

Question 26.
Name the four positively charged sols. .
Answer:
The following are the positively charged sols.

• Hydroated metallic oxide sols. Eg : Al₂O₃.xH₂O, CrO₃.xH₂O etc.
• Basic dye stuffs Eg : Methylene blue sol.
• Hemoglobin (blood).
• Oxides Eg : TiO₂ sol.

Question 27.
Name the four negatively charged sols.
Answer:
The following are the negatively charged sols.

• Metal sols Eg : Ag, Au-sols.
• Metallic suphides sols Eg : ArS₃, CdS sols.
• Acid dye stuffs Eg: eosinol.
• Sols of starch, gum, clay etc.

Question 28.
Explain the terms helmholtz electrical double layer and zeta potential. What are their significancies in the colloidal solutions ?
Answer:
Helmholtz electrical double layer :
The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer.

• In a colloidal sol the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.
• The above concepts applicable in colloidal solution in which the solid particles carry one kind of charge while liquid medium carries opposite charge to that of solids.

Question 29.
Explain with a neat sketch the phenomenon of electrophoresis.
Answer:
Electrophoresis : The existence of charge on colloidal particles is confirmed by electro-phoresis experiment.
When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal articles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Positive charged particles move towards the cathode while negative charged particles towards anode.

Question 30.
Explain the following terms, i) Electrophoresis ii) Coagulation iii) Tyndall effect.
Answer:
Electrophoresis: When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Coagulation : The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If this charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.
Coagulation of lyophobic sols can be carried out by
i) Electrophoresis, ii) Boiling, iii) Adding Electrolytes, iv) Prolonged dialysis etc.

Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

Question 31.
Explain the phenomenon observed
i) When a beam of light is passed through a colloidal sol.
Answer:
Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light. A true solution does not show Tyndall effect.

ii) An electrolyte, NaCl is added to hydrated ferric oxide.
Answer:
When NaCl solution is added to the colloidal solution of hydrated ferric oxide coagulation takes place (precipitation formed).

The reason is that colloidal particles interact with ions (Na⁺, Cl^–) carrying charge opposite to that present on themselves. This causes neutralisation of charges leading to their coagulation.

iii) An electric current is passed through a colloidal solution.
Answer:
Electrophoresis : the existence of charge on colloidal particles is confirmed by electrophoresis experiment.

When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Positive charged particles move towards the cathode while negative charged particles towards anode.

Question 32.
Describe cottrell smoke precipitator with a neat diagram.
Answer:
Aerial pollution by colloidal particles of smoke be prevented by electrical precipitation of smoke. Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust, etc., in air. The smoke, before it comes out from the chimney, is led through a precipitator containing plates having a charge opposite to that carried by smoke particles. The particles on coming in contact with these plates lose their charge and get precipitated. The particles thus settle down on the. floor of the chamber. The precipitator is called conttrels precipitator.

Question 33.
Among NaCl, Na₂SO₄, Na₃PO₄ electrolytes, which is more effective for coagulation of hydrated ferric oxide sol and why ?
Answer:
Among NaCl, Na₂SO₄, Na₃PO₄ electrolytes, Na₃PO₄ is more effective for coagulation of hydrated ferric oxide sol. This is explained by Hardy-Schulze rule.
Greater the valence of the coagulating ion added, the greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Coagulating ability order of the anions in the above salts is
PO₄^-3 > SO₄^-2 > Cl^–.

Question 34.
Discuss how a lyophilic colloids protect a lyophobic colloids.
Answer:
Lyophilic colloids are much stable than lyophobic colloids. A very small amount of electrolyte can precipitate the lyophobic colloids. But their precipitation can be prevented by the addition of small amounts of lyophilic colloids. This lyophilic colloids form a protecting layer around the lyophobic colloids and protect them from precipitation (coagulation). Lyophilic colloids used for. this purpose are called protective colloids.
Ex : Gelatin, gum, starch sols are generally used as protective colloids.

Question 35.
Discuss the use of colloids in
i) Purification of drinking water ii) Tanning iii) Medicines.
Answer:
i) Purification of drinking water: The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

ii) Tanning: Animal skins are colloidal in nature. When a skin, which has positively charged particles, is soaked in tannin, which contains negatively charged colloidal particles, mutual coagulation takes place. This results in the hardening of skin (leather). This process is termed as tanning. Chromium salts are also used in place of tannin.

iii) Medicines : Most of the medicines are colloidal in nature. For example argyrol is a silver sol used as an eye lotion. Colloidal antimony is used in curing kalaazar. Colloidal gold is used as intramuscular injection. Milk of magnesia, an emulsion, is used for stomach disorders. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Question 36.
Define Gold Number.
Answer:
The capacity of the lyophilic colloid in protecting a lyophobic colloid is measured in terms of Gold Number, introduced by “Zigmondy”.

Gold Number:” It is the number of milligrams of lyophilic colloid that is to be added to 10 ml of standard gold sol to prevent its precipitation by the addition of 1ml of a 10% sodium chloride solution”.
Lesser in the gold number the greater is the protecting capacity. Gold number of a few colloids are given below.
Starch = 25 .
Gelatin = 0.005 – 0.01
Albumin =0.1 – 0.2

Question 37.
How do emulsifiers stabilize emulsion ? Name two emulsifiers.
Answer:
Emulsifying agent: The third substance which is added in small amounts to an emulsion to stabilize the emulsion is called emulsifying agent.
—> The emulsifying agent forms an interfacial film between suspended particles and the medium.
Eg : 1) Carbon black stabilizes water in oil type emulsions.
2) Casein and silica stabilize oil in water type emulsions.

Long Answer Questions

Question 1.
Explain the terms absorption, adsorption and sorption. Describe the different types of adsorption.
Answer:
Absorption : The uniform distribution of a substance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Adsorption : The accumulation (or) concentration of a substance on the surface, rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂, etc., on charcoal

Sorption : In case of some substances both adsorption and absorption takes place. This phenomenon is called sorption.

Types of adsorption : On the basis of forces present between adsorbate and adsorbent molecules, adsorption classified into two types :
1) Physisorption, 2) Chemisorption.

1) Physisorption: If the forces that are responsible for the adsorption of adsorbate molecules on the surface of the adsorbent are weak vander waal’s for the adsorption is known as physisorption.
Eg : Adsorption of H₂, on O₂, on charcoal.
Characteristics :

• It is a weak adsorption.
• Enthalpy of adsorption is low (20 – 40 KJ / mole).
• It is reversible and occurs rapidly.
• It decrease with increase in temperature.
• It increase with increase in pressure.
• It is multilayered and not specific.

2) Chemisorption : If the forces that are responsible for the formation of adsorbate molecules on the source of adsorbent are chemical forces, the adsorption is called chemisorption.
Eg : Adsorption of H₂ on Ni – metal surface.
Characteristics :

• It is a strong adsorption. ,
• Enthalpy of adsorption of high (80 – 240 KJ / mole).
• It is irreversible and occurs slowly.
• It increases with increase of temperature and finally decrease.
• Pressure has no effect on chemisorption.
• It is unilayered and specific.

Question 2.
Discuss the characteristics of physical adsorption.
Answer:
Characteristics of physical adsorption :

• Lack of specificity : Physisorption is not specific. In this adsorption the given surface of adsorbent doesnot show any preference for a particular gas as the forces present are variderwaal’s forces (Universal).
• Nature of adsorbate : The amount of gas adsorbed by a solid depends on the nature of gas. Easily liquefiable gases are readily adsorbed.
• The gas with high critical temperature value is easily liquefied and gets readily adsorbed.
• Reversible Nature : Physical adsorption is reversible and occurs rapidly at low temperatures.
• Surface area of adsorbent: The extent of adsorption increase with increase in surface area of adsorbent. .
• Finely divided metals and porous substances having large surface area are good adsorbents.
• Enthalpy of adsorption : Physisorption is an exothermic process but the enthalpy of adsorption is low (20 – 40 KJ/mole). This is due to the presence of Vander waal’s forces.

Question 3.
Discuss the characteristics of chemisorption. .
Answer:

1. High specificity: Chemisorption is highly specific and it will only occur if there is some possibility of chemical bonding between adsorbate and adsorbent.
2. Irreversibility : Chemisorption involves compound formation and is irreversible in nature.
   • Chemisorption is exothermic and increases with increase of temperature.
3. Surface area : Chemisorption increase with increase of surface area of the adsorbent.
4. Enthalpy of adsorption : Enthalpy of chemisorption is high (80 – 240 KJ/mole).

Question 4.
Compare and contrast the phenomenon of physisorption and chemisorption.
Answer:
Characterstics of Physical adsorption :
Physisorption: If the forces that are responsible for the adsorption of adsorbate molecules on the surface of the adsorbent are weak vander waal’s for the adsorption is known as physisorption.
Eg : Adsorption of H₂ on O₂ on charcoal.

• Lack of specificity : Physisorption is not specific. In this adsorption the given surface of adsorbent doesnot show any preference for a particular gas as the forces present are vander waal’s forces (Universal). ‘
• Nature of adsorbate : The amount of gas adsorbed by a solid depends on the nature of gas.
• Easily liquefiable gases are readily adsorbed:
  The gas with high critical temperature value is easily liquefied apd gets readily adsorbed.
• Reversible Nature : Physical adsorption is reversible and occurs rapidly at low temperatures.
• Surface area of adsorbent: The extent of adsorption increase with increase in surface area of adsorbent.
• Finely divided metals and porous substances having large surface area are good adsorbents. .
• Enthalpy of adsorption : Physisorption is an exothermic process but the enthalpy of adsorption is low (20 – 40 KJ/mole). This is due to the presence of Vander waal’s forces.

2) Chemisorption : If the forces that are responsible for the formation of adsorbate molecules on the source of adsorbent are chemical forces, the adsorption is called chemisorption.
Eg : Adsorption of H₂ on Ni – metal surface.

1. High specificity: Chemisorption is highly specific and it will only occur if there is some possibility of chemical bonding between adsorbate and adsorbent.
2.  Irreversibility : Chemisorption involves compound formation and is irreversible in nature.
   • Chemisorption is exothermic and increases with increase of temperature.. ,
3. Surface area : Chemisorption increase with increase of surface area of the adsorbent.
4. Enthalpy of adsorption : Enthalpy of chemisorption is high (80 – 240 KJ/mole).

Question 5.
What is an adsorption isotherm ? Discuss the phenomenon of adsorption of gases on solids with the help of Freundlich adsorption isotherm.
Answer:
Adsorption Isotherm : The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve known as adsorption Isotherm.

• Freundlich adsorption isotherm equation is
  \(\frac{\mathrm{x}}{\mathrm{m}}\) = k, P^1/n
  x = mass of the gas adsorbed
  m = mass of the adsorbent
  P, k and n are constants.
• k and n depend on the nature of the adsorbent and the gas at a particular temperature.
• The relation ship is generally represented in the form of a curve where x/m is plotted against pressure.
• These curves indicate that at a fixed pressure there is a decrease in physical adsorption with rise of temperature.
  
  \(\frac{\mathrm{x}}{\mathrm{m}}\) = k.P^1/4
  Applying logarithm
  Log \(\frac{\mathrm{x}}{\mathrm{m}}\) = log k + \(\frac{\mathrm{1}}{\mathrm{n}}\) log p.
• The validity of Freundlich isotherm can be verified by plotting log \(\frac{\mathrm{x}}{\mathrm{m}}\) on y-axis and log P on x-axis. If it is a straight line then the isotherm is valid.
• The slope of the straight line give the value of \(\frac{\mathrm{1}}{\mathrm{n}}\)
• The intercept on the y-axis gives the value of log k.

Question 6.
Give a detailed account of applications of adsorption.
Answer:
Applications of adsorption : The phenomenon of adsorption funds a number of applications. Important ones are listed here.

1. Production of high vacuum: The traces of air still remaining in the vessel evacuated by a vacuum pump to give high vacuum can be adsorbed by charcoal.
2. Gas masks : Gas mask, a device which consists of activated charcoal or mixture of adsorbents is usually used by coal miners to adsorb poisonous gases during breathing.
3. Control of humidity : Silica gel and alumina gel are used as adsorbents for removing mixture and controlling humidity of air in rooms.
4. Removal of colouring matter from solutions : Animal charcoal removes colours of impure coloured solutions by adsorbing impurities responsible for the colour.
5. Separation of inert gases : Due to the difference in degree of adsorption of gases by charcoal, a mixture of noble gases can be separated by adsorption on coconut charcoal at different temperatures.
6. In curing diseases : A number of drugs kill germs by getting themselves adsorbed on germs.
7. Froth floatation process : A low grade sulphide ore is concentrated by separating silica and other earthy matter by this method using pine oil and frothing agent.
8. Adsorption indicators : Surfaces of certain precipitates such as silver halides have the property of adsorbing some dyes like eosin, fluorescein, etc. and thereby producing a characteristic colour change at the end point in argentometric titrations.
9. Chromatographic analysis : Chromatographic analysis based on the phenomenon of adsorption finds a number of applications in analytical and industrial methods.

Question 7.
What is catalysis ? How is catalysis classified ?* Give four examples for each type of catalysis.
Answer:
Catalysis: A substance which alters the rate of a chemical reaction without itself being consumed in the process, is called a catalyst.
The action of catalyst in altering the rate of a chemical reaction is called catalysis.
Types of catalysis : Catalysis is classified into two types as .
a) Homogeneous catalysis and
b) Heterogeneous catalysis.
Homogeneous catalysis : The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

Heterogeneous catalysis: The catalytic process in which the catalyst is present in a phase different from that of reactarts is known as heterogeneous catalysis.

Question 8.
Discuss the mechanism of heterogeneous catalysis.
Answer:
Adsorption theory of heterogeneous catalysis :
Adsorption theory explains the mechanism of heterogeneous catalysis.

• The modern adsorption theory is the combination of the intermediate compound formation theory and the old adsorption theory.
• The mechanism of catalysis involves five steps.
• Diffusion of the reactants to the surface of the catalyst.
• Adsorption of the reactant molecules on the surface of the catalyst.
• Occurrence of chemical reaction on the catalyst’s surface through formation of an intermediate.
• Desorption of reaction products from the catalyst surface, and thereby making the surface available again for more reaction to occur.
• Diffusion of reaction products away from the catalyst’s surface.

Question 9.
What are enzymes ? Explain in detail the enzyme catalysis, with necessary examples.
Answer:
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.

• These act as specific catalysts in biological reactions.
• These catalyse the numerous reactions that occur in the bodies of animals and plants to maintain the life process.
  Mechanism of enzyme catalysis :
  
  The enzyme – catalysed reactions may be considered to proceed in two steps.
  Step 1 : Binding of substrate to enzyme to form an activated complex (ES^≠)
  E + S → ES^≠
  Step 2 : Decomposition of the activated complex to form product.
  ES^≠ → E + P

Factors influencing the catalytic activity of enzymes :

• Highly specific nature: One catalyst catalyses only one reaction. Each enzyme is specific for a given reaction.
  Eg : The enzyme urease catalyses the hydrolysis of urea only and not any other amide
• Highly active under optinum temperature: The rate of an enzyme reaction is maximum at a temperature called optimum temperature.
• The optimum temperature range for enzymatic activity is 298 – 310 K.
•  Highly active under optimum pH : The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH which lies between 5-7
•  Increasing activity in presence of activators and co-enzymes : The enzymatic activity is increased in presence of co-enzymes (or) in presence of activators such as Na⁺, Mn⁺² etc.,
• Influence of inhibitors and poisons: The inhibitors (or) poisons interact with the active functional groups on the enzyme surface and reduce or completely destroy the catalytic activity.
  i) Inversion of Cane sugar: Enzyme : Invertase
  

Question 10 & 11.
What are colloidal solutions ? How are they classified ? Give examples.
Answer:
A heterogeneous system in which one substance is dispered as large particles in another substance is called colloidal solution.

• In colloidal solution the particle size of the dispersed phase is of the order of 1 mμ – 1 μ where as in true solution the particle size of the solute are the order of mp or less.
• Colloidal solution is a heterogeneous binary system where as true solution is a homogeneous binary system.
  On the basis of physical states of dispersed phase and dispersion medium colloidal solutions are classified into eight types.
  

On the basis of nature of dispersion medium colloids are classified as follows.
If the dispersion medium is ‘air’, the sols are called aerosols.
Eg : Smoke.

If the dispersion medium is ‘water’ the sols are called Hydrosol (or) aquasol.
Eg: starch solution.
If the dispersion medium is ‘alcohol’ then the sols are called alcosols.

The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution. Ex : Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 12.
How are colloids classified on the basis of the nature of interaction between a dispersed phase and a dispersion medium ? Describe an important characteristic of each class. Which of the sols need stabilising agents for preservation ?
Answer:
The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.

The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Lyophilic colloids are much stable than lyophobic colloids. Avery small amount of electrolyte can precipitate the lyophobic colloids. But their precipitation can be prevented by the addition of small amounts of lyophilic colloids. This lyophilic colloids form a protecting layer around the lyophobic colloids and protect them from precipitation (coagulation). Lyophilic colloids used for this purpose are called protective colloids.
Ex : Gelatin, gum, starch sols are generally used as protective colloids.

Question 13.
What are micelles ? Discuss the mechanism of micelle formation and cleaning action of soap.
Answer:
Micelles : Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.
Eg : Stearate ions (C₁₇H₃₅COO^–) associate together in high concentration, in a solution of soap in water and forms a micelle.

Mechanism of micelle formation : Let us take the example of soap solution. Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOC^– Na⁺ (E.g., sodium strearate CH₃ (CH₂)₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na⁺ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ’tail) which is hydrophobic (water repelling), and a polar group COO^–(also called polar-ionic ‘head), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. But at critical micelle concentration, the COO^– irons are pulled into the bulk of the solution and are aggregated to form a spherical shape with their hydro¬carbon chains pointing towards the centre of the sphere with COO- part remaining outward on the surface of the sphere. The aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 simple molecules.

Similarly, in case of detergents, E.g., sodium laurylsulphate. CH₃ (CH₃)₁₁ SO₃^–Na⁺, the polar group is -SO₃^–along with the long hydrocarbon chain. Hence, the mechanism of micelle formation here also is same as that of soaps.
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.
C₁₇H₃₅COONa → C₁₇H₃₅COO^– + Na⁺ (Stearate ion)
The anion of soap = C₁₇H₃₅COO^–

Cleaning action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

Soap thus, functions as an emulsifying agent for the water dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 14.
Describe the properties of colloids with necessary diagrams wherever necessary.
Answ:
i) Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles.. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light.
A true solution does not show Tyndall effect.

ii) Brownion movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig-zag motion.
This rapid motion of colloidal particles is called Brownian movement.

This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

iii) Charge on colloidal particles : The charge on the colloidal particles is due to electron capture by sol particles during electro dispersion of metals and due to preferrential adsorption ions from solution (of) due to formulation of electrical double layer.

When a dilute AgNO₃ solution is added to a dilute KI solution taken in excess the precipitated AgI adsorbs I^– ions present in excess and a negatively charged AgI colloidal solution is formed. However when dilute KI solution is added to dilute AgNO₃ solution taken in excess, AgI precipitate adsorbs Ag⁺ ions present in excess and a positively charged Agl colloidal solution is formed. Generally the ion common to dispersed particle is adsorbed.
AgI/I^–
Negatively charged
AgI/Ag⁺
Positively charged.

iv) Zeta Potential: In a colloidal sol the charges of opposite signs on the fixed’and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.

v) electrophoresis : When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in opposite direction. This phenomenon is termed “electro osmosis”.

vi) Coagulation : The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If tins charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.

Question 15.
What are emulsions ? How are they classified ? Describe the applications of emulsion. [T.S. Mar. 18, 16]
Answer:
Emulsion: The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called Emulsion.
Ex: Milk. .
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.
Classification of emulsions : Emulsions are classified, into two classes. These are
a) oil in water (O/W) and ‘
b) water in oil (W/O), (O = Oil; W = Water).
These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.

a) Oil in Water (O/W) type emulsions : In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex : Milk, liquid, fat (oil) in water.
Vanishing cream; fat in water.

b) Water in Oil (W/O) type emulsions : In this type of emulsions, the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil
Cold cream : water in fat.

Applications of Emulsions : Emulsions are useful

• In the digestion of fats in intestines.
•  In washing processes of clothes and crockery.
•  In the preparation of lotions, creams, ointments in pharmaceutical and cosmetics.
• In the extraction of metals (froth flotation).
• In the conversion of cream into butter by churning.
• To break oil and water emulsions in oil wells.
• In the preparation of oily type of drugs for easy adsorption to the body.

Intext Questions

Question 1.
Write any two characteristics of Chemisorption. .
Solution:

1. High specificity: Chemisorption; is highly specific and it will only occur when adsorbent and adsorbate molecules can chemically react with each other.
   Eg.: oxygen is adsorbed on metals by oxide formation.
2. Surface area : Chemisorption increases with increase in surface area of the adsorbent.

Question 2.
Why does physisorption decrease with the increase of temperature ?
Solution:
Physisorption is an exothermic process :
Solid (Adsorbent) + Gas (Adsorbate) ⇌ Gas adsorbed on solid + Heat
When temperature is increased, the equilibrium shifts towards the backward direction to neutralise the effect of the change (Le-Chatelier’s principle). So, physisorption decreases with increase in temperature.

Question 3.
Why are finely powdered substances more effective adsorbents than their non powdered crystal forms ? .
Solution:
Finally divided substances provide increased surface area for adsorption which is not available to such extent in their crystalline forms. That’s why the powdered forms are more effective than crystalline forms for the purpose of adsorbents.

Question 4.
Hydrogen used in Haber’s process is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming. Why is it necessary to remove CO formed in steam reforming when ammonia is obtained by Haber’s process ?
Solution:
Carbon monoxide (CO) acts as a poison for catalyst iron and promoter molybdenum in Haber’s process, i.e., the efficiency of catalyst and promoter is decreased. It also combines with Fe to form iron carbonyl, Fe(CO)₅ which interfere in the production of ammonia. Hence, CO must be removed from the reaction mixture.

Question 5.
Why is ester hydrolysis slow in the beginning but is fast after sometime ?
Solution:
The chemical equation for ester hydrolysis is as follows :

Carboxylic acid produced during hydrolysis releases H+ ions in the solution which act as catalyst (auto-catalysis) for the reaction. Therefore, ester hydrolysis is slow in the beginning but becomes faster after sometime.

Question 6.
What is role of desorption in the process of adsorption catalysis.
Solution:
The reaction products formed on the catalyst surface get detached from the surface as a result of desorption, thereby making the surface available again for more reactant particles.

Question 7.
What modification can you suggest in the Hardy-Schulze law ?
Solution:
According to Hardy Schulze law, the ions carrying charge opposite to the charge on sol particles neutralise their charge and thus cause their coagulation or precipitation. But actually, the sol carrying these ions also get coagulated since the ions neutralize their charge. So, Hardy Schulze law can be modified as :

“When equimolar proportions of two oppositely charged sols are mixed, they mutually neutralize their charge and both get coagulated”.

Question 8.
Why is it essential to wash the precipitate in gravimetric chemical analysis with wash liquid before drying and weighing it quantitatively ?
Solution:
Some of the reactant ions my be adsorbed or may adhere to the surface of the particles of the precipitate formed during an ionic reaction. In order to remove these reactant ions, the precipitate should.be washed with water. If this is not done, an error may be produced during quantitative analysis.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption

Very Short Answer Questions

Question 1.
Give the dental formula of an adult human being.
Answer:
The arrangement of different types of teeth in each half of both the jaws in the order I, C, PM, M is represented by the dental formula.
In adult it is = \(\frac{2123}{2123}\) = 32.

Question 2.
Bile juice contains no digestive enzymes, yet it is important for digestion. How?
Answer:
Bile juice doesn’t contain enzymes, but it contains bile salts such as sodium/potassium glycocholates and taurocholates, which help in the digestion and absorption of lipids. Bile salts emulsify fats and also render them water-soluble. Bile salts activate the lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

Question 3.
Describe the role of chymotrypsin. Name two other digestive enzymes of the same category secreted by the same gland.
Answer:
Chymotrypsin plays an important role in digestion of proteins, proteoses and peptones and convert them into tripeptides and dipeptides. Chymotrypsin, trypsin and carboxy peptidase are ehdopeptidases, produced by the pancreas arid involved in the digestion of proteins.

Question 4.
What would happen if, HCl were not secreted in the stomach?
Answer:
HCl is secreted by the glands present on the stomach walls. It provides acidic pH which is optimal for the action of pepsin. HCl activates the pepsinogen into pepsin. Pepsin plays an important role in digestion of proteins. Therefore, if HCl were not secreted in the stomach, then pepsin would not be activated. This would affect on protein digestion.

Question 5.
Explain the terms thecodont and diphyodont dentitions.
Answer:
Thecodont:
Teeth of human beings are embedded in the sockets of the jaw bones is called thecodont.

Diphyodont:
Majority of mammals including human beings form two sets of teeth during their life time, a set of temporary / milk teeth replaced by a set of permanent teeth. This type of dentition is called diphyodont dentition.

Question 6.
What is Autocatalysis? Give two examples.
Answer:
Autocatalysis is the catalysis of a reaction in which the catalyst is one of the product of the reaction (or) catalysis caused by a catalytic agent formed during a chemical reaction.

Question 7.
What is chyme?
Answer:
Semi fluid mass of partly digested acidic food formed in the stomach is called chyme.

Question 8.
Name the different types of Salivary glands of man, and their locations in the human body?
Answer:
There are three pairs of Salivary glands in man.
1. Parotid glands — present below the pinna / inner surface of the cheeks.
2. Sub maxillary (or) Sub mandibular glands —.located at the angles of lower jaw.
3. Sublingual glands — present below the tongue.

Question 9.
Name different types of papillae present on the tongue of man.
Answer:
The upper surface of the tongue has small projections called papillae. In humans the tongue bears 3 (three) types of papillae namely 1) fungi form 2) filiform 3) Circumvallate papillae.

Question 10.
What is the hardest substance in the human body? What is its origin?
Answer:
Enamel of tooth is the hardest substance in the human body, which is secreted by ameloblasts of ectodermal origin.

Question 11.
Name the structure of gut which is vestigial in human beings, but well developed in herbivores. And mention the type of tissue with which it is mostly formed.
Answer:
Appendix is vestigial part in human beings. It is a narrow finger like tubular projection, arises from the caecum. In herbivores it is a functional part and useful in the digestion of cellulose materials. The appendix contain a high concentration of lymphoid tissues. These are highly specialized structures which are a part of the immune system.

Question 12.
Distinguish between deglutition and mastication.
Answer:
Deglutition :
Deglutition is the swallowing of food and involves a complex and coordinated process. It is divided into three phases.

Phase one :
The collection and swallowing of masticated food.

Phase two :
Passage of food through the pharynx into the beginning of the esophagus.

Phase three :
The passage of food into the stomach.

Mastication :
The mastication process includes the biting and tearing of food into manageable pieces. This usually involves using the incisors and canines teeth. The grinding of food is usually performed by the molars and premolars. During the mastication process, food is moistened and mixed with saliva.

Question 13.
Distinguish between diarrhoea and constipation.
Answer:
Diarrhoea :
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. It reduces the absorption of fqod arid results in loss of water.

Constipation :
A condition ip which the faeces are retained within the rectum as it is hard due to low content of water and the movement of bowel occurs irregularly.

Question 14.
Name two hormones secreted by the duodenal mucosa.
Answer:
The epithelium of duodenum secretes the hormones namely secretin and cholecystokinin (cck).

Question 15.
Distinguish between absorption and assimilation.
Answer:
Absorption :
Absorption is the process by which the end products of digestion pass through the intestinal mucosa into blood (or) lymph. It is carried out by passive, active (or) facilitated transport mechanisms.

Assimilation:
The absorbed substances finally reach the tissues, where food materials become integral components of the living protoplasm and used for the production of energy, growth and repair. This process is called assimilation.

Short Answer Questions

Question 1.
Draw a neat labelled diagram of L.S of a tooth. Ans.
Answer:

Question 2.
Describe the process of digestion of proteins in the stomach.
Answer:
Protein digestion begins in the stomach. The food entered into the stomach is mixed thoroughly with the gastric juice of the stomach by the churning movements of its muscular wall and the product is called chyme. The main components of gastric juice are protein digestive enzymes, hydrochloric acid and mucus.

HCl provides the acidic pH (1.8) which is optimal for the action of pepsin. The proenzymes of gastric juice, the pepsinogen and prorennin, on exposure to hydrochloric acid are convened into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin is a proteolytic enzyme found in the gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions and converts it into calcium paracaseinate and proteoses. Pepsin acts on calcium paracaseinate and converts it into peptones. The entire process of protein digestion in the stomach takes about 4 hours.

Question 3.
Explain the role ofpancreatic juice in the digestion of proteins.
Answer:
Pancreatic juice is secreted by the pancreas and it plays an important role in protein ‘digestion. Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsinogen and pro carboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.

Chymotrypsin, Trypsin and Carboxy Peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides. Which in turn hydrolysed into aminoacids by the action Of tri aruj di peptidases.

Question 4.
How are polysaccharides and disaccharides digested?
Answer:
Dietary carbohydrates principally consist of polysaccharides :
Starch and glycogen. It contains disaccharides and small amounts of monosaccharides.

Digestion in mouth:
Digestion of carbohydrates starts at the mouth, where they come in contact with saliva during mastication. Saliva contains carbohydrate-splitting enzyme called Salivay amylase (ptyalin). This enzyme hydrolyses the starch into disaccharides (maltose).

Digestion in stomach:
Ptyalin action stops in stomach when pH falls to 3.0. No carbohydrate splitting enzymes are available in gastric juice. Some dietary sucrose may be hydrolysed by HCl.

Digestion in small intestine :
Chyme reaches the duodenum from stomach where it meets pancreatic juice. Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosaccharides.

Question 5.
If you take butter in your food how does it get digested and absorbed in the body? Explain.
Answer:
Butter contains fat. Fats remain mostly undigested in stomach.

Digestion of fat in the small intestine :
The major site of fat digestion is the small intestine. This is due to the presence of a powerful lipase/(steapsin) in the pancreatic juice and bile juice. Bile juice contains bile salts such as Sodium/Potassium glycocholates and taurocholates, which helps in the emulsification of fat i.e., break down of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

Absorption :
Fatty acids and glycerol being insoluble in water cannot be absorbed into the blood directly. They are first modified into small droplets called micelles, which move into intestinal mucosal cells. They are reformed into very small protein coated fat globules called chylomicrons, which are transported into the lymph capillaries in the villi by exocytosis. Then they are ultimately released into blood stream through left subclavian vein via the thoracic duct. These chylomicrons are broken down to fatty acids and glycerol by the action of an enzyme lipoprotein lipase and they diffuse into the adipocytes of the adipose tissue and liver for storage.

Question 6.
What are the functions of liver?
Answer:
Liver performs a variety of functions such as synthesis, storage and secretion of various substances.

1. Liver secretes bile juice, it contains bile salts such as sodium / potassium glycocholates and taurocholates, which helps in digestion and absorption of lipids.
2. Liver plays the key role in carbohydrate metabolism.
   a) Glycogenesis : formation of glycogen from glucose.
   b) Glycogenolysis : breakdown of glycogen into glucose.
   c) Gluconeogenesis : Synthesis of glucose from certain amino acids, lactate (or) glycerol.
3. Liver also plays an important role in synthesis of cholesterol and production of triglycerides.
4. Deamination of proteins occurs in the liver.
5. Liver is the chief organ of detoxification of toxic substances that enter the gut along with food.
6. Liver acts as thermoregulatory organ.
7. Liver acts as a haemopoietic organ in the foetus and erythroclastic organ in the adult.
8. The liver synthesizes the plasma proteins such as albumin, globulins, blood clotting factors such as fibrinogen / prothrombin, etc., and the anticoagulant called heparin.
9. The lactic acid formed during anaerobic muscle contraction is converted into glycogen (gluconeogenesis) in the liver by Cori cycle.
10. Kupffer cells are the largest phagocytic cells which remove unwanted substances and microbes that attack the liver by phagocytosis.

Long Answer Questions

Question 1.
Describe the physiology of digestion of various types of food in the human digestive system
Answer:
Digestion is the process of convertion of complex non-diffusible food substances into simple diffusible forms. The process of digestion is accomplished by mechanical and biochemical process.

I. Digestion in the buccal cavity :
Buccal cavity performs two major functions, mastication of food and facilitation of swallowing. Teeth and tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into bolus. The saliva secreted into oral cavity contains electrolytes such as Na⁺, K⁺, Cl^–, HCO₃^– arid enzymes like salivary amylase (ptyalin) and lysozyme. Carbohydrates digestion starts in the buccal cavity, about 30% of starch is hydrolysed here into a disaccharide called maltose by the enzyme amylase (ptyaline). Lysozyme acts as antibacterial agent that prevents infections.

II. Digestion in the stomach :
As the bolus enters into stomach starch digestion stops and protein digestion begins. The food entered into stomach is mixed thoroughly with gastric juice of the stomach by the churning movement of its muscular wall and the product is called chyme. The mucus and bicarbonates present in the gastric juice act as lubricant and protect the mucosal epithelium from HCl. HCl in the stomach provides the acidic pH (1.8) which is optimal for the action of pepsin.

The proenzymes of gastric juice, the pepsinogen and prorennin on exposure to HCl are converted into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin found in gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions convert into calcium paracaseinate and proteoses. Pepsin acts on paracaseinate and convert it into peptones. The entire process of protein digestion in stomach takes about 4 hours.

III. Digestion in the small intestine :
Various types of movements are generated by the muscular external layer of small intestine. These movements help in thorough mixing of the food with bile, pancreatic juice and intestinal juice in the intestine and thereby facilitate digestion. The duodenal cells of the proximal part produces large amount of bicarbonates to completely neutralize any gastric acid that passes further down into the digestive tract.

i) Digestion of proteins :
Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsin and procarboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase secreted by the intestinal mucosa into active trypsin which intum activate the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.

Chymotrypsin, trypsin and carboxy peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides which inturn hydrolysed into amino acids by the action of tri and dipeptidases.

ii) Digestion of fats:
Bile salts of bile help in the emulsification of fat i.e., breakdown of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

iii) Digestion of Carbohydrates :
Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosacharides.

iv) Digestion of nucleic acids :
Nucleases of the pancreatic juice act on the nucleic acids to form nucleotides and nucleosides. Nucleotidases and nucleosidases of the intestinal juice convert the nucleotides and nucleosides into pentose sugar and nitrogen bases.

The end products of digestion pass through the intestinal mucosa into blood (or) lymph is carriedout by passive, active (or) facilitated transport mechanisms.

Question 2.
Explain the digestive system of man with neat labelled diagram.
Answer:

The digestive system is a group of organs and tissues involve in the breaking down of ingested food in the alimentary canal into a form that can be absorbed ai a assimilated by the tissues of the body.

Human digestive system consists of the alimentary canal and the associated glands. Alimentary canal / Digestive tract:

The alimentary canal of man begins with the anterior opening, the mouth and ends with the posterior opening, the anus.

Parts of the alimentary canal / digestive tract:

1. Mouth and Buccal (oral) cavity
2. Pharynx
3. Oesophagus
4. Stomach
5. Small intestine
6. Large intestine

1. Mouth and Buccal (oral) cavity:
Mouth is the first part of the alimentary canal. It is formed by the cheek on either side and boardered by the movable upper and lower lips, leads into the buccal (or) oral cavity. The palate separate the ventral buccal cavity from the dorsal nasal chamber and facilitates chewing and breathing simultaneously. The jaw bones bear teeth and tongue occurs at the base of the buccal cavity.

i) Teeth :
These are ecto-mesodermal in origin. An adult human has 32 permanent teeth, which are of four different types namely, incisors (I), canines (C), premolars (PM), and molars (M). These are useful in cutting, tearing and grinding of food. The arrangement’ of teeth is represented by dental formula. In adult human
it is = \(\frac{2123}{2123}\) = 32

ii) Tongue :
It is a freeely movable muscular sense organ, attached to the floor of the oral cavity by a fold of tissue called frenulum. The upper surface of the tongue has small projections called papillae, some of which bear taste buds. The tongue acts as universal toothbrush and helps in mixing saliva with food, taste detection, deglutition and speaking.

2. Pharynx:
It is a muscular tube connecting the oral cavity and oesophagus and trachea. It is a common passage for food and air. It is divided into nasopharynx, oropharynx and laryngo pharynx. Oesophagus and trachea open into the laryngopharynx. The trachea open into the laryngopharynx through the glottis. A cartilaginous flap called epiglottis prevents the entry of food into glottis during swallowing.

3. Oesophagus:
It is a thin long muscular tube (9 to 12 inches). The semisolid digested food from pharynx enters the oesophagus. Oesophagus is separated by the Cardiac sphincter from stomach. When the food reaches lower end of Oesophagus the cardiac sphincter opens allowing the food to enter the stomach.

4. Stomach :
It is a wide ‘J’ shaped muscular sac, located iii the upper left portion of the abdominal cavity just below the diaphragm. It has three major parts, an anterior cardiac portion into which the oesophagus opens, a middle large fundic region and a posterior pyloric portion which opens into the first part of the small intestine through the pyloric aperture which is guarded by the pyloric sphincter.

5. Small intestine :
The small intestine is the longest part of alimentary canal. It has three regions namely proximal duodenum middle long coiled jejunum and distal highly coiled ileum. Duodenum receives the hepato-pancreatic duct. Ileum opens into the large intestine.

6. Large intestine :
It consist of caecum, colon and rectum. Caecum is a small blind sac. A narrow finger like vestigial tubular organ arises from caecum called appendix. The caecum opens into colon which is‘divided into an ascending, a transverse, a descending parts and a sigmoid colon that continues behind into rectum. Rectum is a small dilated sac which leads into anal canal that opens out through the anus.

Digestive glands :
1. Salivary glands : There are three pairs of glands in man.
i) Parotid glands
ii) Sub-maxillary glands
iii) Sub-lingual glands
They secrete saliva, which mainly contains salivary amylase and lysozyme.

2. Gastric glands :
These are located in the wall of the stomach beneath the surface
epithelium, Gastric glands are of three types namely
i) Cardiac glands – secrete mucus
ii) Pyloric glands – secrete mucus and hormone gastrin
iii) Fundic / Oxyntic glands – secrete mucus, proenzymes like pepsinogen and prorennin, HCl, intrensic factor and some amount of gastric lipase.

3. Intestinal glands :
They are of two types
i) Brunner’s glands
ii) Crypts of lieberkuhn
which secrete intestinal juice contains peptidases, disaccharidases, enterokinase and lysozyme.

4. Liver :
Liver is the largest gland in human. Liver secretes bile juice, contains bile salts, which play a very important role in lipid digestion.

5. Pancreas :
The pancreas is the second largest gland in human. Exocrine part of pancreas secretes pancreas juice contains sodium bicarbonates, trypsinogen, chymotrypsinogen, carboxy peptidase, steapsin, -pancreatic amylase and nucleases such as DNAase and RNAase.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(c) Group-17 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(c) Group-17 Elements

Very Short Answer Questions

Question 1.
Interhalogen compounds are more reactive than the constituent halogens except fluorine – Explain.
Answer:
Inter halogen compounds are more reactive than halogens. This is because X – X’ bond in interhalogens is weaker than X — X bond in halogens except F – F bond.

Question 2.
What is the use of ClF₃?
Answer:
ClF₃ is very useful fluorinating agent and it is used for the production of VF₆ in the enrichment of U²³⁵.
U_(s) + 3ClF_3(l) → UF_6(g) + 3ClF_(g)

Question 3.
Why are halogens coloured?
Anšwer:
Halogens are coloured due to the absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. Halogens absorb different quanta of radiation and display different colours.

Question 4.
Write the reactions of F₂ and Cl₂ with water. (TS Mar. 17) (IPE Mar. ‘14)
Answer:
Fluorine produces O₂ and O₃ on passing through water.
3F₂ + 3H₂O → 6HF + O₃
2F₂ + 2H₂O → 4HF + O₂

Question 5.
Electron gain enthalpy of fluorine is less than that of chlorine – explain.
Answer:
The negative electron gain enthalpy of the fluorine is less than that of chlorine. This is due to small size of fluorine atom which results in strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and thus the incoming electron does not experience much attraction.

Question 6.
Bond dissociation enthalpy of F₂ is less than that of Cl₂ – Explain.
Answer:
Bond dissociation enthalpy of F₂ is less than that of Cl₂
Explanation:
In F₂ molecule electron repulsions are greater among lone pairs because these lone pairs are much closer to each other than in case of Cl₂.

Question 7.
Write the formulae of the compounds, in which oxygen has positive oxidation states and mention the oxidation states of oxygen in them.
Answer:
In OF₂ and O₂F₂ oxygen has positive oxidation states.

• In OF₂, oxygen oxidation state is + 2.
• In O₂F₂ oxygen oxidation state is + 1.

Question 8.
What is the use of O₂F₂ and I₂O₅?
Answer:
Uses of O₂F₂

• O₂F₂ is a fluorinating agent. O₂F₂ oxidises plutonium to PUF₆ and the reaction is used in removing plutonium as PUF₆ from spent nuclear fuel.

Use of I₂O₅:

• I₂O₅ is a good oxidising agent and is used in the estimation of carbon monoxide (CO).

Question 9.
Explain the reactions of Cl₂ with NaOH. (IPE – 2015 (AP), 2016 (TS), (AP))
Answer:
i) Reaction with cold dilute NaOH: Chlorine reacts with cold dilute NaOH to give sodium hypochlorite and sodium chloride.

ii) Reaction with hot concentrated NaOH : Chlorine reacts with hot concentrated NaOH to give sodium chlorate and sodium chloride.

Question 10.
What happens when Cl₂ reacts with dry slaked lime ? (AP Mar. 17: IPE 16, 15 (AP))
Answer:
Chlorine reacts with dry slaked lime and forms bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.

Question 11.
What is aqua regia ? Write its reaction with gold and platinum.
Answer:
A mixture of 3 parts of conc. HCl and one part of Conc. HNO₃ constitutes aqùa regia. It is used for dissolving noble metals.
It’s reaction with gold:
Au + 4H⁺ + \(\mathrm{NO}_3^{-}\) + 4Cl^– → \(\mathrm{AuCl}_4^{-}\) + NO + 2 H₂O
It’s reaction with Platinum:
3Pt + 16H⁺ + \(4 \mathrm{NO}_3^{-}\) + 18Cl^– → 3PtC\(l_6^{-2}\) + 4NO + 8 H₂O

Question 12.
How is chlorine manufactured by Deacon’s method? (AP Mar.’ 17: IPE ’16 (TS))
Answer:
Deacon’s process: [n Deacon’s process. Chlorine is obtained by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ (catalyst) at 723 K.

Question 13.
Why is dry chlorine cannot act as a bleaching agent.
Answer:
Dry chlorine cannot produce nascent oxygen. Hence it cannot act as a bleaching agent.

Question 14.
HF is a liquid while HCl is a gas — explain.
Answer:
HF is a liquid due to the presence of inter molecular hydrogen bonding where as HCl is a gas as there is no such type of bonding in it.

Question 15.
Write two uses of hydrogen chloride.
Answer:
Uses of hydrogen chloride:

• It is used in medicines and as a laboratory reagent.
• It is used in the manufacturing of Cl₂, NH₄Cl and glucose (from corn starch).
• It is used in extracting glue from bones and purifying bone black.

Question 16.
Chlorine acts as an oxidizing agent – explain with two examples.
Answer:
Chlorine acts as oxidising agent.
Example – 1: Cl₂ oxidises Iodine to Iodate.
I₂ + 6 H₂O + 5 Cl₂ → 2HIO₃ + 10 HCl

Example – 2: Cl₂ oxidises Sodium Sulphite to Sodium Sulphate.
Cl₂ + Na₂SO₃ + H₂O → Na₂SO₄ + 2 HCl

Question 17.
Write the reaction of chlorine with hypo (Na₂S₂O₃). (IPE Mar.2015 (AP, TS), 2016 (TS), (AP))
Answer:
Reaction of chlorine with hypo (Na₂S₂O₃). Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2HCl + S. In this reaction hypo acts an “antichlor”.

Question 18.
Give the bond dissociation order of halogens.
Answer:
Bond dissociation order of halogens is, Cl₂ > Br₂ > F₂ > I₂.

Question 19.
I₂ is more soluble in KI give reason.
Answer:
I₂ forms soluble KI₃ with KI solution. I₂ + KI → KI₃.

Question 20.
Chlorine acts as a bleaching agent only in the presence of moisture — explain.
Answer:
Moist chlorine is a powerful bleaching agent. This bleaching property is due to oxidation.
Cl₂ + H₂O → 2HCl + (O)
Ex: Coloured substance + (0) → colourless substance.

Question 21.
The decreasing order of acidic character among hypohalogen acids is HClO > HBrO > HIO. Give reason.
Answer:
Given the decreasing order of acidic character among hypohalogen acids is
HClO > HBrO > HIO

From the above mentioned Ka values the given order of hypohalogen acids – acid strength order is
HClO > HBrO > HIO

Question 22.
fluorine exhibits only-1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Explain.
Solution:
Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Other halogens have d- orbitais and therefore, can expand their octets and show +1, +3, +5 and +7 oxidation states also.

Question 23.
What are the inter halogen compounds ? Give two examples.
Answer:
The compounds formed between different halogens are called inter halogen compounds.
Ex: ClF₃, BrF₃, IF₇, ICl₃.

Short Answer Questions

Question 1.
Explain the structures of
a) BrF₅ and
b) IF₇.
Answer:
a) Structure of BrF₅.

• Central atom in BrF₅ is ‘Br’
• ‘Br’ undergoes sp³d² hybridisation in 2^nd excited state.
  
• Shape of the molecule is octahedral with one position occupied by a lone pair (or) square pyramidal.
  

b) Structure of IF₇:

• Central atom in IF₇ is 1’.
• ‘I’ undergoes sp³d³ hybridisation iñ 3^rd excited state.
  
• Shape of the molecule is Pentagonal bipyramid structure.
  

Question 2.
How is chlorine obtained in the laboratory? How does it react with the following?
a) cold dil. NaOH
b) excess NH₃
c) KI (IPE Mar & May – 2015 AP TS), BMP)
Answer:
In the laboratory chlorine is prepared by the oxidation of HCl with MnO₂.
4 HCl + MnO₂ → MnCl + 2 H₂O + Cl₂↑
a) Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
b) Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chloride are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂↑
c) Cl₂ reacts with KI to liberate iodine.
Cl₂ + 2KI → 2KCl + I₂↑

Question 3.
How is ClF₃ prepared? How does it react with water? Explain its structure.
Answer:
Preparation of ClF₃: Chlorine reacts with excess of fluorine to form ClF₃.

Reaction with H₂O : ClF₃ reacts with water explosively and oxidises water to give oxygen or in controlled quantitieš oxygen diflouride (OF₂) as well as HF and HCl.
ClF₃ + 2 H₂O → 3 HF + HCl + O₂
ClF₃ + H₂O → HF + HCl + OF₂

Structure of ClF₃:

• Central atom in ClF₃ is ‘Cl’
• Excited state electronic configuration of ‘Cl’ is
  
• Cl’ atom undergoes sp³d hybridisation.
• It is a bent T-shaped molecule (or) trigonal bipyramidal with 2 – positions occupied by lone pairs.
  

Question 4.
How can you prepare Cl₂ from HCl and HCl from Cl₂ ? Write the reactions.
Answer:
Preparation of Cl₂ from HCl:

• On heating MnO₂ with conc. HCl, Cl₂ gas is liberated.
  MnO₂ + 4 HCl → MnCl₂ + Cl₂ + 2 H₂O
• By the oxidation of HCl gas by atmospheric oxygen in presence of CuCl₂ catalyst at 723K.
  
  Preparation of HCl from Cl₂:
• Cl₂ when reacted with H₂ to form HCl,
  H_2(g) + Cl_2(g) → 2HCl_(g)

Question 5.
How is chlorine prepared by electrolytic method? Explain Its reaction with
a) NaOH and
b) NH₃ under different conditions. (IPE Mar & May – 2015 (AP, TS))
Answer:
Preparation of chlorine by electrolytic method: Chlorine is obtained by the electrolysis of brine solutions (Cone. NaCl). Cl₂ gas is liberated at anode.
2 NaCl → 2 Na⁺ + 2Cl^–
2 H₂O + 2e^– → 2 OH^– + H₂ (cathode)
2 Cl^– → Cl₂ + 2e^– (anode) .

a)

i)NaOH:
Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
ii) Cl₂ reacts with hot conc. NaOH to form sodium chloride and sodium chlorate.
3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

b)

i) Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chlorate are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂↑
ii) NH₃ reacts with excess of Cl₂ to form NCl₃ and HCl.

Question 6.
What are interhalogen compounds? Give some examples to illustrate the definition. How are they classified?
Answer:
The binary diamagnetic compounds of halogens which are formed by the reaction of halogens among themselves are called interhalogen compounds.
E.g.: IF₇, ClF₃, BrF₃, ClF, IF₃ etc.
The above examples are binary diamagnetic compounds and formed by combination of halogens only.
Inter halogen compounds are classified into four types.

1. AX – Týpe :Eg: ClF, BrF
2. AX₃ – Tÿpe : Eg : ClF₃, IF₃
3. AX₅ – Type : Eg: ClF₅, BrF₅
4. AX₇ – Type : Eg : IF₇
   • ‘A’ is less electronegative halogen.
   • X is more electronegative halogens.

Question 7.
Write the names and formulae of the oxoacids of chlorine. Explain their structures and relative acidic nature.
Answer:
Four oxyacids of chlorine are known. They are
Hypochlorous acid – HOCl
Chlorous acid — HClO₂
Chloric acid – HClO₃
Perchloric acid – HClO₄
Structure of HClO: In this chlorine atom is sp³ hybridised. Outer electronic configuration of Cl in ClO^– after sp³ hybridisation.

Shape is tetrahedral with 3 lone pairs (or) linear
No π — bonds.
Chlorous acid: (HClO₂) : Chlorine is in sp³ hybrid state, in first excited state. Shape is tetrahedral with 2 lone pairs (or) angular one π_d-p bond is present.
First excited state

Chloric acid (HClO₃) : The central chlorine atom undergoes sp³ hybridisation in second excited state.
Second excited state

Shape is tetrahedral with one lone pair (or) pyramidal.
Two π_d-p bonds present.

Perchloric acid (HClO₄) : The central chlorine atom undergoes sp³ hybridisation in third excited state.

Shape is perfect tetrahedral. No lone pairs.
Three π_d-p bonds present.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers

Very Short Answer Questions

Question 1.
Explain why propanol has a higher boiling point than that hydrocarbon-butane.
Answer:
Propanol has a higher boiling point (391 K) than that hydrocarbon butane (309K).

Reason: In propanol, strong intermolecular hydrogen bonding is present between the molecules. But in the case of butane weak van der Waals force of attractions are present.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explanation: .

• Alcohols and water are both polar solvents. Alcohol is dissolves in water, due to formation of hydrogen bonding with water molecules.
• Hydro carbons are non polar and these dorv.t form hydrogen bonds with water molecules. So alcohols are soluble in water where as hydrocarbons are not soluble in water.

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:
Given molecular formula of monnhydric phenols is C₇H₈O. The no. of possible isomers with molecular formula C₇H₈O are three.

Question 4.
Give the reagents used for the preparation of phenol from chiorobenzene.
Answer:
Phenol is prepared from chiorobenzene as follows. Reagents required are

1. NaOH, 623K, 300 atm
2. HCl.
   Chemical reaction :
   

Question 5.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
Only 1° – alcohols form Ethers on acid dehydration. But not 2° or 3°-alcohols.

Reason : In case of 2° or 3° alcohols steric hindrance arises. Due to this steric hindrance alkenes are formed but not Ethers.

Question 6.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
Case – I: When methoxy methane reacts with cold.dil. HI then methyl alcohol and methyl iodide are formed.
Mechanisms:

Case – II : When methoxy methane reacts with hot.conc.HI then only methyl iodide is formed.
Mechanisms:

Question 7.
Name the reagents used in the following reactions.

1. Oxidation of primary alcohol to carboxylic acid
2. Oxidation of primary alcohol to aldehyde.

Answer:

1. The reagents used for the oxidation of 1° – alcohols to carboxylicacid are acidified K₂Cr₂O₇7 (or) Acidic/alkaline KMnO₄ (or) Neutral KMnO₄
2. The reagents used for the oxidation of 1°- alcohols to aldehyde are pyridine chloro chromate (PCC) in CH₂Cl₂.

Question 8.
Write the equations for the following reactions.
i) Bromination of phenol to 2,4, 6-tribromophenol
ii) Benzyl alcohol to benzoic acid.
Answer:
i) Bromination of phenól to 2, 4, 6 tribromophenol.

Question 9.
IdentIfy the reactant needed to form t-.butylalcohol from acetone.
Answer:
When acetone reacts with methyl magnesium bromide followed by the hydrolysis forms t-butyl alcohol.

Question 10.
Write the structures for the following compounds

1. Ethoxyethane
2. Ethoxybutane
3. Phenoxyethane

Answer:

1. Ethoxyethane → CH₃ – CH₂ – O – CH₂ – CH₃
2. Ethoxybutane → CH₃ – CH₂ – O – CH₂ – CH₂ – CH₂ – CH₃
3. Phenoxyethane → C₆H₅ – O – CH₂ – CH₃

Short Answer Questions

Question 1.
Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O₂ and give their IUFAC names and classify them as primary, secondary and tertiary alcohols.
Answer:

• Given molecular formula of compound is C₅H₁₂O.
• It has eight isomeric alcohols.
  
  

In the above isomeric alcohols (i), (ii), (iii), (iv) and 1°-alcohols; (v), (vi) and (viii) are 2°- alcohols, (vii) is 3°-alcohol.

Question 2.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give rèason.
Answer:
While separating a mixture of ortho and para nitrophenols by steam distillation, ortho nitrophenol is steam volatile.

Reason: In ortho nitrophenol intra molecular hydrogen bonding is present and in case of para nitrophenol inter molecular hydrogen bonding is present. So O-nitrophenol is steam volatile.

Question 3.
Give the equations for the preparation of phenol from Cumene. [Mar. 14]
Answer:
Phenol.is prepared from Cumene as follows.

1. Oxidation of Cumene to Cumene hydroperoxide.
2. Cumene hydroperoxide on acidic hydrolysis to form phenol.
   

Question 4.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Hydration of Ethene to yield ethanol involves 3—step mechanism.
Step – 1: In step – 1 formation of carbocation takes place by the protonation of ethene.

Step – 2 : In step – 2 carbocation formed in the above step attacked by water.

Step – 3 : Ethyl alcohol is formed by he deprotonation in step -3

Question 5.
Explain the acidic nature of phenols and compare with that of alcohols.
Answer:
The reaction of phenol with sodium metal and with aq.NaOH indicates the acidic nature of phenol.
i) Phenol reacts with sodium metal to form sodium phenoxide.

•  In phenol hydroxyl group is attached to the Sp² hydridised carbon of benzene ring which acts as electron with drawing group. The formed phenoxide ion from phenol is more stabilised due to delocalisation of negative charge.

Comparison of acidic character of Phenol and Ethanol:

• The reaction of phenol with aq. NaOH indicates that phenols are stronger acids than alcohols.
• The hydroxyl group attached to an aromatic ring is more acidic than in hydroxyl group is attached to an alkyl group.
• Phenol forms stable phenoxide ion stabilised by resonance but ethoxide ion is not.
  

Question 6.
Write the products formed by the reduction and oxidation of phenol.
Answer:
a) Reduction of phenol: Phenol undergo reduction in presence of zinc dust to form benzene.

b) Oxidation of phenol : Phenol undergo oxidation with chromicacid and forms a conjugated diketone known as benzoquinone.

Question 7.
Ethanol with H₂SO₄, at 443K forms ethene while at 413 K it forms ethoxy ethane. Explain the mechanism.
Answer:
Case – 1: Ethanol reacts with Cone. H₂SO₄ at 443K forms ethene

Mechanism:
Step – 1: Formation of protonated alcohol

Step 2 : Formation of carbo cation

Step 3 : Formation of ethene by elimination of a proton

Case – II : Ethanol reacts with Cone. H₂SO₄ at 413 K to form ethoxy ethane.

Mechanism:
In the above reaction ether formation is a SN reaction. This involve attack of alcohol molecule on a protonated alcohol.

Question 8.
Account for the statement: Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.
Answer:
Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.

Explanation : Consider ethanol, propane and methoxy methane which are having comparable molecular masses. The boiling points, molecular masses and structures of the above compounds mentioned below.

The higher boiling points of alcohols are due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.

Question 9.
Explain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences + R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophihic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m—position. So 0-and P-products are mainly formed during electrophillic substitution reactions.

Question 10.
Write the products of the following reactions :

Answer:

Long Answer Questions

Question 1.
Write the IUPAC name of the following compounds :

Answer:

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
i) 2, Methyl butan—ol
ii) 1-Phenylprpan-2-ol
iii) 3, 5-Dhuethylhexane-1, 3, 5-triol
iv) 2, 3-Diethylphenol
v) 1-Ethoxypropane
vi) 2-Ethoxy-3-methylpentane
vii) Cyclohexylmethanol
viii) 3-Chloromethylpentan-1-ol
Answer:

Question 3.
Write the equations for the preparation of phenol using benzene, conc. H₂SO₄ and NaOH. [Mar. 14]
Answer:
The equations for the preparation of phenol using conc.H₂SO₄ and NaoH as follows

Question 4.
Illustrate hydroboration-oxidation reaction with a suitable example.
Answer:
When alkenes undergo addition reaction with diborane to form tri alkyl boranes. These followed by the oxidation by alkaline H₂O₂ to form alcohols. This reaction is called as hydroboration-oxidation reaction.

Question 5.
Write the IUPAC name of the following compounds:

Answer:

Question 6.
How will you synthesise:
i) 1 – Phenylethanol from a suitable alkene
ii) Cyclohexylmethanol using an alkyl halide by an SN² reaction.
iii) Pentan-1-ol using a suitable alkyl halide.
Answer:
i) Synthesis of 1-phenylethanol from a suitable alkene : When styrene undergo hydrolysis in presence of dil.H₂SO₄ to form 1-phenyl ethanol. It is an example of Marknowni koff’s rule.

ii) Synthesis of cyclohexyl methanol using an alkyl halide by SN² reaction : When cyclohexyl methyl bromide reacts with aq. NaOH to form cyclohexyl methanol.

iii) Synthesis of 1-pentanol using a suitable alkyl halide: When 1-Bromo pentane reacts with aq.KOH to form 1-pentanol.

Question 7.
Explain why-
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
ii) OH group attached to benzene ring activates it towards electrophilic substitution.
Answer:
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
Explanation: .

• -NO₂ is an electron withdrawing group and -OCH₃ is electron releasing group.
• By the presence of electron withdrawing group the phenoxide ion is more stabilized. By the presence of electron releasing group the phenoxide ion is less stabilized.
• Due to high stability of phenoxide ion, acidic nature increases.
  

ii) The -OH group attached to benzene ring activates it towards electrophilic substitution.

Explanation : When an electrophile is attacked, the – OH group exerts +R effect on the benzene ring. So electrodensity in the ring increases at ortho and para positions. When an electrophile attacks, substitution takes place at O and p-positions. So benzene ring activates towards electrophilic substitution.

Question 8.
Wth a suitable example write equations for the following:. [T.S. MAr. 19, 18; A.P. Mar. 18, 16] [A.P. Mar. 16]
i) Kolbe’s reaction
ii) Reimer-Tiemann reaction
iii) Williamsons ether synthesis
Answer:
i) Kolbes reaction: Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophilhic substitution with CO₂to form salicylic acid.

ii) Relmer-Tlemann reactIon : Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzaldehyde). This reaction is known as Reimer-Tiemann reaction. .

iii) Wilhiamsons ether synthesis:

• This method is used for the preparation of symmetrical and unsymmetrical ethers.
• The reaction of an alkyl halide with sodium alkoxide to form ethers is known as Williamsons Synthesis.
  

Question 9.
How are the following conversions carried out?
i) Benzyl chloride to Benzyl alcohol
ii) Ethyl magnesium bromide to Propan-1-ol
iii) 2-Butanone to 2-Butanol
Answer:
i) Conversion of Benzyl chloride to Benzyl alcohol : Ben.zyl chloride reacts with aq. NaOH to form benzyl alcohol.

ii) Conversion of Ethyl magnesium bromide to Propan-1-ol : Ethyl magnesium bromide reacts with form aldehyde followed by hydrolysis to form 1-propanol.

iii) Conversion of 2-Butanone to 2-Butanol: 2-Butanone undergo reduction in presence of LiA/H₄ to form 2-Butanol.

Question 10.
Write the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
i) 1-Propoxypropane
ii) Ethoxybenzene
iii) 2-Methoxy-2-methylpropane
iv) 1 -Methoxyethane
Answer:
i) Preparation of 1-propoxy propane :

Question 11.
How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.
Answer:
1 – Proponal reacts with Conc. H₂SO₄ at 413 K to form 1-propoxy propane.

Question 12.
Explain the fact that in aryl alkyl ethers the alkoxy group activates the benzene ring towards electrophilic substitution.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences +R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophillic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m-position. So O-and P-products are mainly formed during electrophillic substitution reactions.

Question 13.
Write equations of the below given reactions:
i) Alkylation of anisole
ii) Nitration of anisole
iii) Friedel-Crafts acetylation of anisole
Answer:
i) Friedel crafts Alkylation of anisole :

ii) Nitration of anisole

Question 14.
Show how you would synthesize the following alcohols from appropriate alkenes?

Answer:

Question 15.
Explain why phenol with bromine water forms 2,4,6-tribromophenol while on reaction with bromine in CS₂ at low temperatures forms para-bromophenol as the major product.
Answer:
a) Phenol under goes Bromination in presence of CS₂ to form p-bromophenol as major product.

b) Phenol undergoes bromination in aqueous medium form 2,4,6 -tribromo phenol (white ppt).

Explanation: In bromination of phenol, the polarisation of Br₂ molecule takes place even in the absence of Lewis acid. This is due to the highly activating effect of -OH group attached to the benzene ring.

Textual Examples

Question 1.
Give IUPAC names of the following compounds:

Solution:
i) 4-Chloro-2, 3-dimethylpentan-1-ol
ii) 2-Ethoxypropane
iii) 2, 6-Dimethyiphenol
iv) 1-Ethoxy-2-nitrocyclohexane

Question 2.
Give the structures and IUPAC names of the products expected from the following reactions :
a) Catalytic reduction of butanal.
b) Hydration of propene in the presence of dilute sulphuric acid.
c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis.
Solution:

Question 3.
Arrange the following sets of compounds in order of their increasing boiling points :
a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
Solution:
a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.

Question 4.
Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3, 5-dinitrophenol, phenol, 4-methylphenol.
Solution:
Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 5.
Write the structures of the major products expected from the following reactions:
a) Mononitratlon of 3-methylphenol .
b) Dinitratlon of 3methylphenol
c) Mononitration of phenyl methanoate.
Solution:
The combined influence of -OH and -CH₃ groups determine the position of the incoming group.

Question 6.
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

i) What would be the major product of this reaction?
ii) Write a suitable reaction for the preparation of t-butylethyl ether.
Solution:
i) The major product of the given reaction is 2-methylprop-1-ene.
It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

Question 7.
Give the major products that are formed by heating each of the following ethers with HI.

Solution:

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:

Answer:
Primary alcohols (i), (ii), (iii)
Secondary alcohols (iv) and (y)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols are (ii) and (vi)

Question 3.
Name the following compounds according to IUPAC system.


Answer:
i) 3-Chioremethyl 2-isopropylpentan-1-ol
ii) 2, 5-Dimethylhexane-1, 3-dial
iii) 3-Bromocyclohexanol
iv) Hex-1-en-3-ol
v) 2-Bromo-3-methylbut-2-en-1-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent of methanol?

Answer:

Question 5.
Write structures of the products of the following reactions :


Answer:

Question 6.
Predict the major product of acid catalysed dehydration of

1. 1-methylcyclohexanol and
2. butan-1-ol.

Answer:

1. 1-Methylcyclohexene
2. A mixture of but-1-ene and but-2-ene. But-1-ene is the major product formed due to rearrangement to give,secondary carbocation.

Question 7.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Question 8.
Predict the products of the following reactions:

Answer:

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 2 System of Circles to solve questions creatively.

Intermediate 2nd Year Maths 2B System of Circles Formulas

Definition:
→ The angle between two intersecting circles is defined as the angle between the tangents at the point of intersection of the two circle’s. If θ is the angle between the circles, then
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
Here d = distance between the centres, r₁, r₂ be their radii.

→ If θ is angle between the two circles.
x² + y² + 2gx + 2fy + c = 0 and x² + y² + 2g’x + 2f’y + c’ = 0, then
cos θ = \(\frac{-2 g g^{\prime}-2 f f^{\prime}+c+c^{\prime}}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

→ Circles cut orthogonally if
2g’g + 2f’f = c + c’ [∵ cos 90° = 0]
(or) d² = r²₁ + r²₂ then also two circles cut orthogonally.

Theorem:
If d is the distance between the centers of two intersecting circles with radii r₁, r₂ and θ is the angle between the circles then cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\).

Proof:
Let C₁, C₂ be the centre s of the two circles S = 0, S’ = 0 with radii r₁, r₂ respectively. Thus C₁C₂ = d. Let P be a point of intersection of the two circles. Let PB, PA be the tangents of the circles S = 0, S’ = 0 respectively at P.

Now PC₁ = r₁, PC₂ = r₂, ∠APB = θ
Since PB is a tangent to the circle S = 0, ∠C₁PB = π/2
Since PA is a tangent to the circle S’ = 0, ∠C₂PA = π/2
Now ∠C₁PC₂ = ∠C₁PB + ∠C₂PA – ∠APB = π/2 + π/2 – θ = π – θ
From ∆C₁PC₂, by cosine rule,
C₁²C₂² = PC₁² + PC₂² – 2PC₁ . PC₂ cos ∠C₁PC₂ ⇒ d² = r₁² + r₂² – 2r₁r₂ cos(π – θ) ⇒ d² = r₁² + r₂² + 2r₁r₂ cos θ
⇒ 2r₁r₂ cos θ = d² – r₁² – r₂² ⇒ cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\)

Corollary:
If θ is the angle between the circles x² + y² + 2gx + 2fy + c = 0, x² + y² + 2g’x + 2f’y + c’= 0 then cos θ = \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
proof:
Let C₁, C₂ be the centre s and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively and C₁C₂ = d.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’),
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c}\)
Now cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\) = \(\frac{\left(g-g^{\prime}\right)^{2}+\left(f-f^{\prime}\right)^{2}-\left(g^{2}+f^{2}-c\right)-\left(g^{\prime 2}+f^{\prime 2}-c^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
= \(\frac{\mathrm{g}^{2}+\mathrm{g}^{\prime 2}-2 \mathrm{gg}^{\prime}+\mathrm{f}^{2}+\mathrm{f}^{\prime 2}-2 \mathrm{ff} \mathrm{f}^{\prime}-\mathrm{g}^{2}-\mathrm{f}^{2}+\mathrm{c}-\mathrm{g}^{2}-\mathrm{f}^{\prime 2}+\mathrm{c}^{\prime}}{2 \sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}} \sqrt{\mathrm{g}^{2}+\mathrm{f}^{\prime 2}-\mathrm{c}^{\prime}}}\)
= \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

Note: Let d be the distance between the centers of two intersecting circles with radii r₁, r₂. The two circles cut orthogonally if d² = r₁² + r₂².
Note: The condition that the two circles

S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’ = 0 may cut each other orthogonally is 2gg’ + 2ff’ = c + c’.
Proof: Let C₁, C₂ be the centers and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’)
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c^{\prime}}\)
Let P be point of intersection of the circles.
The two circles cut orthogonally at P
⇔ ∠C₁PC₂ = 90°.
⇒ C₁C₂² = C₁P² + C₂P² ⇔ (g – g’)² + (f – f’)² = r₁² + r₂²;
⇔ g² + g’² – 2gg’ + f² + f’² – 2ff’ = g² + f² – c + g’² + f’² + c’
⇔ – (2gg’ + 2ff’) = – (c + c’) ⇒ 2gg’+ 2ff’ = c + c’

Note:

• The equation of the common chord of the intersecting circles s = 0 and s¹ = 0 is s – s¹ = 0.
• The equation of the common tangent of the touching circles s = 0 and s¹ = 0 is s – s¹ = 0
• If the circle s = 0 and the line L = 0 are intersecting then the equation of the circle passing through the points of intersection of s = 0 and L = 0 is S + λL = 0.
• The equation of the circle passing through the point of intersection of S = 0 and S’ = 0 is s + λS’ = 0.

Theorem: The equation of the radical axis of the circles S = 0, S’ = 0 is S – S’ = 0.

Theorem: The radical axis of two circles is perpendicular to their line of centers.
Proof:
Let S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’= 0 be the given circles.

The equation of the radical axis is S – S’ = 0
⇒ 2(g – g’)x + 2(f – f’)y + (c – c’) = 0
⇒ a₁x + b₁y + c₁ = 0 where
a₁ = 2(g – g’), b₁ = 2(f – f’), c₁ = e – e’
The centers of the circles are (- g, – f), (- g’, – f’)
The equation to the line of centers is:
(x + g) (f – f’) = (y + f) (g – g’)
⇒ (f – f’)x – (g – g’)y – gf’ + fg’= 0
⇒ a₂x + b₂y + c₂ = 0 where
a₂ = f – f’, b₂ = – (g – g’), c₂ = fg’ – gf’
Now a₁a₂ + b₁b₂ = 2(g – g’) (f – f’) – 2(f – f’) (g – g’) = 0.

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B System of Circles Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B System of Circles Important Questions

Question 1.
x² + y² + 4x + 8 = 0, x² + y² – 16y + k = 0 [A.P. & T.S. Mar. 16]
Solution:
g₁ = 2; f₁ = 0; c₁ = 8
g₂ = 0; f₂ = – 4; c₂ = k
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(2) (0) + 2(0) (-8) = 8 + k
0 + 0 = 8 + k
⇒ k = -8

Question 2.
(x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c² (a ≠ b) [A.P. Mar. 15]
Solution:
(x² + y² – 2xa – 2yb – c²)
– (x² + y² – 2xb – 2ya – c²) = 0
– 2x (a – b) – 2y(b – a) = 0
(or) x – y = 0

Question 3.
Find the angle between the circles given by the equations. [T.S. Mar.17]
i) x² + y² – 12x – 6y + 41 = 0, x² + y² + 4x + 6y – 59 = 0.
Solution:
C₁ = (6, 3)
r₁ =(36 + 9 – 41)^1/2
r₁ = 2

C₂ = (-2, -3)
r₂ =(4 + 9 + 59)^1/2
r₂ = (72)^1/2 = 6\(\sqrt{2}\)

C₁C₂ = d = \(\sqrt{(6+2)^{2}+(3+3)^{2}}\)
= \(\sqrt{64+36}\) = 10
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
= \(\frac{100-4-72}{2 \times 2 \cdot \sqrt{72}}=\frac{24}{4 \times 6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
θ = 45°

Question 4.
Find the equation of the circle which cuts the circles x² + y² – 4x – 6y + 11 = 0 and x² + y² – 10x – 4y + 21 = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7. [A.P. Mar. 16; May 07]
Solution:
Let circle be x² + y² + 2gx + 2fy + c = 0 ……………….. (i)
Orthogonal to circle
2g (-2) +2f(-3) = 11 + c ……………………. (ii)
2g (-5) + 2f(-2) = 21 + c ……………………. (iii)
Subtracting it we get
-6g + 2f = 10 ……………………… (iv)
Circles centre is on 2x + 3y = 7
∴ -2g – 3f = 7 ……………………. (v)
Solving (iv) and (y)
f = -1, g = -2, c = 3
Equation of circle be x² + y² – 4x – 2y + 3 = 0

Question 5.
Show that the angle between the circles x² + y² = a², x² + y² = ax + ay is \(\frac{3\pi}{4}\) [Mar. 14]
Solution:
Equations of the circles are
S ≡ x² + y² – a² = 0
S’ ≡ x² + y² – ax – ay = 0
C₁ (0, 0), C₂ (\(\frac{a}{2}\), \(\frac{a}{2}\))
∴ C₁C₂² = (0 – \(\frac{a}{2}\))² + (0 – \(\frac{a}{2}\))²

Question 6.
If x + y = 3 ¡s the equation of the chord AB of the circle x² + y² – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter. [A.P. Mar. 15]
Solution:
Required equation of circle passing through
intersection S = 0 and L = 0 is S + λL = 0
(x² + y² – 2x + 4y – 8) + λ(x + y – 3) = 0
(x² + y² + x(-2 + λ)+ y(4 + λ) – 8 – 3λ = 0 ………………. (i)
x² + y² + 2gx + 2fy + c = 0 ……………………. (ii)
Comparing (i) and (ii) we get
g = \(\frac{(-2+\lambda)}{2}\), f = \(\frac{(4+\lambda)}{2}\)
Centre lies on x + y = 3
∴ \(-\left(\frac{-2+\lambda}{2}\right)-\left(\frac{4+\lambda}{2}\right)\) = 3
2 – λ – 4 – λ = 6
-2λ = 8 ⇒ λ = – 4
Required equation of circle be
(x² + y² – 2x + 4y – 8) – 4(x + y – 3) = 0
x² + y² – 6x + 4 = 0

Question 7.
If two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’. [T.S. Mar. 16]
Solution:
C₁ = (-g, -f)
r₁ = \(\sqrt{g^{2}+f^{2}}\)
C₂ = (-g^-1, -f^-1)
r₂ = \(\sqrt{g^{\prime 2}+r^{\prime 2}}\)
C₁C₂ = r₁ + r₂
(C₁C₂)² = (r₁r₂)²
(g’ – g)² + (f’ – f)² = g² + f² + g’² + f’² + \(2 \sqrt{g^{2}+f^{2}} \sqrt{g^{2}+f^{1^{2}}}\)
-2(gg’ + ff’) = 2{g²g’² + g²f’² + f²g’²}^1/2
Squaring again
(gg’ + ff’)² = g²g’² + f²f’² + g²f’² + g’²f²
g²g’² + f²f’² + 2gg’ff’ = g²g’² + f²f’² + g²f’² + g’²f’²
2gg’ff’ = g²f’² + f²g’²
⇒ g²f’² + g’²f’² – 2gg’ff’ = 0
(or) (gf’ – fg’)² = 0 (or) gf’ = fg’

Question 8.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 …………………… (1)
and S ≡ x² + y² + 6x + 2y – 90 = 0 …………………(2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10.
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In
this case, the common tangent is nothing but the radical axis. Therefore its equation is S – S’ = 0.
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) in the ratio 5 : 10
i.e., 1: 2 (externally)
∴ Point of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5)

Question 9.
Find the angle between the circles x² + y² + 4x – 14y + 28 = 0 and x² + y² + 4x – 5 = 0
Solution:
Equations of the given circles are .
x² + y² + 4x – 14y + 28 = 0
x² + y² + 4x – 5 = 0
Centres are C₁ (-2, 7), C₂ (-2, 0)
C₁C₂ = \(\sqrt{(-2+2)^{2}+(7-0)^{2}}\)
= \(\sqrt{0+49}\) = 7
r₁ = \(\sqrt{4+49-28}\) = \(\sqrt{25}\) = 5
r₂ = \(\sqrt{4+5}\) = \(\sqrt{9}\) = 3
If θ is the angle between the given circles,
then cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
cos θ = \(\frac{49-25-9}{2(5)(3)}\) = \(\frac{15}{2.5 .3}\) = \(\frac{1}{2}\) = cos 60°
Angle between the circles = θ = \(\frac{\pi}{3}\)

Question 10.
If the angle between the circles x² + y² – 12x – 6y + 41 = 0 and x² + y² + kx + 6y – 59 = 0 is 45° find k.
Solution:
Suppose θ is the angle between the circles
x² + y² – 12x – 6y + 41 = 0
and x² + y² + kx + 6y – 59 = 0
g₁ = -6, f₁ = -3, c₁ = 41,
g₂ = \(\frac{k}{2}\), f₂ = 3, c₂ = -59

\(\frac{1}{\sqrt{2}}=\frac{6 k}{4 \cdot \sqrt{\frac{k^{2}}{4}+68}}\)
Squaring and cross – multiplying
4(\(\frac{k^{2}}{4}\) + 68) = 18k²
\(\frac{2\left[k^{2}+272\right]}{4}\) = 9k²
k² + 272 = 18 k²
17k² = 272
k² = \(\frac{272}{17}\)
k² = 16
k = ±4.

Question 11.
Find the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles.
x² + y² – 8x – 2y + 16 = 0 and …………….. (1)
x² + y² – 4x – 4y – 1 = 0. ………………. (2)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………. (3)
Then the circle (3) is orthogonal to (1) and (2).
∴ By applying the condition of orthogonality give in x² + y² + 2gx + 2fy + c = 0 we get
2g(-4) + 2f(-1) = c + 16 and …………………. (4)
2g (-2) + 2f(-2) = c – 1 ……………….. (5).
Given that the circle (3) is passing through (1, 1)
∴ 1² + 1² + 2g(1) + 2f(1) + c = 0
2g + 2f + c + 2 = 0 ………………….. (6)
Solving (4), (5) and (6) for g, f and c, we get
g = –\(\frac{7}{3}\), f = \(\frac{23}{6}\), c = -5
Thus the equation of the required circle is
3(x² + y²) – 14x + 23y – 15 = 0.

Question 12.
Find the equation of the circle which is orthogonal to each of the following three circles.
x² + y² + 2x + 17y + 4 = 0 ………………… (1)
x² + y² + 7x + 6y + 11 = 0 ………………. (2)
and x² + y² – x + 22y + 3 = 0 ………………… (3)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………… (4)
Since this circle is orthogonal to (1), (2) and (3). by applying the condition of orthogonality given in x² + y² + 2gx + 2fy + c = 0
we have
2(g)(1) + 2(f) (\(\frac{17}{2}\)) = c + 4 ……………….. (5)
2(g) (\(\frac{7}{2}\)) + 2(f)(3) = c + 11 ………………. (6)
and 2(g) (-\(\frac{1}{2}\)) + 2(f)(11) = c + 3 ………………….. (7)
Solving (5), (6) and (7) for g, f, c we get
g = -3, f = -2 and c = -44
Thus the equation of the required circle is
x² + y² – 6x – 4y – 44 = 0.

Question 13.
If the straight line is represented by
x cos α + y sin α = p ………………….. (1)
intersects the circle
x² + y² = a² ……………… (2)
at the points A and B, then show that the equation of the circle with \(\overline{\mathrm{AB}}\) as diameter is(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.
Solution:
The equation of the circle passing through the points A and B is S ≡ x² + y² + 2gx + 2fy + c = 0
(x² + y² – a²) + λ(x cos α + y sin α – p) = 0 ……………… (3)
The centre of this circle is
\(\left(-\frac{\lambda \cos \alpha}{2},-\frac{\lambda \sin \alpha}{2}\right)\)
If the circle given by (3) has \(\overline{\mathrm{AB}}\) as diameter then the centre of it must lie on (1)
∴ \(-\frac{\lambda \cos \alpha}{2}\) (cos α) – \(\frac{\lambda \sin \alpha}{2}\) (sin α) = p
i.e., –\(\frac{\lambda}{2}\) (cos² α + sin² α) p
i.e., λ = -2p
Hence the equation of the required circle is
(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.

Question 14.
Find the equation of the circle passing through the points of intersection of the circles.
x² + y² – 8x – 6y + 21 = 0 ……………….. (1)
x² + y² – 2x – 15 = 0 ……………….. (2)
and (1, 2).
Solution:
The equation of circle passing through the points of intersection of (1) arid (2) is
(x² + y² – 8x – 6y + 21) + λ (x² + y² – 2x – 15) = 0 ……………….. (3)
If it passes through (1, 2). we obtain
(1 + 4 – 8 – 12 + 21) + λ(1 + 4 – 2 – 15) = 0
i.e., 6 + λ(-12) = 0
i.e., λ = \(\frac{1}{2}\)
Hence the equation of the required circle is
(x² + y² – 8x – 6y + 21) + \(\frac{1}{2}\) (x² + y² – 2x – 15) = 0
i.e., 3(x² + y²) – 18x – 12y + 27 = 0.

Question 15.
Let us find the equation the radical axis of the circles S ≡ x² + y² – 5x + 6y + 12 = 0 and S’ ≡ x² + y² + 6x – 4y – 14 = 0
Solution:
The given equations of circles are in general form. Therefore their radical axis is (S – S’ = 0)
i.e., 11x – 10y – 26 = 0

Question 16.
Let us find the equation of the radical axis of the circles
2x² + 2y² + 3x + 6y – 5 = 0 ………………….. (1)
and 3x² + 3y² – 7x + 8y – 11 = 0 ……………….. (2)
Solution:
Hence the given equations are not in general form, we get :
x² + y² + \(\frac{3}{2}\)x + 3y – \(\frac{5}{2}\) = 0 and
x² + y² – \(\frac{7}{3}\)x + \(\frac{8}{3}\) y – \(\frac{11}{3}\) = 0
Now the radical axis equation of given circles is
(\(\frac{3}{2}\) + \(\frac{7}{3}\))x + (3 – \(\frac{8}{3}\))y + (-\(\frac{5}{2}\) + \(\frac{11}{3}\)) = 0
i.e., 23x + 2y + 7 = 0

Question 17.
Let us find the radical centre of the circles
x² + y² – 2x + 6y = 0 ………………… (1)
x² + y² – 4x – 2y + 6 = 0 …………………. (2)
and x² + y² – 12x + 2y + 3 = 0 ………………. (3)
Solution:
The radical axis of (1) and (2) and (3)
x + 4y – 3 = 0 ………………… (4)
8x – 4y + 3 = 0 …………………. (5)
10x + 4y – 3 = 0 ……………… (6)
Solving (4) and (5) for the point of intersection we get (0, \(\frac{3}{4}\)) which is the required radical centre. Observe that the co-ordinates of this point satisfies (6) also.

Question 18.
Find the equation and length of the common chord of the two circles
S ≡ x² + y² + 3x + 5y + 4 = 0
and S’ ≡ x² + y² + 5x + 3y + 4 = 0
Solution:
Equations of the given circles are
S ≡ x² + y² + 3x + 5y + 4 = 0 ………………… (1)
S’ ≡ x² + y² + 5x + 3y + 4 ………………. (2)
Equations of the common chord is S – S’ = 0
-2x + 2y = 0
L ≡ x – y = 0 …………….. (3)

Question 19.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 ……………….. (1)
and S’ ≡ x² + y² + 6x + 2y – 90 = 0 ……………. (2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In this case, the.common tangent is nothing but the radical axis. Therefore its equation is
S – S’ = 0
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides in the ratio 5 : 10 i.e., 1 : 2 (externally)
∴ Point, of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5).

Question 20.
Find the equation of the circle whose diameter is the common chord of the circles
S ≡ x² + y² + 2x + 3y + 1 = 0 ……………….. (1)
and S’ ≡ x² + y² + 4x + 3y + 2 = 0 ……………….. (2)
Solution:
Here the common chord is the radical axis of (1) and (2). The equation of the radical axis is S – S’ = 0.
i.e., 2x + 1 = 0 …………………… (3)
The equation of any circle passing through
the points of intersection of (1) and (3) is (S + λL = 0)
(x² + y² + 2x + 3y + 1) + λ(2x + 1) = 0
x² + y² + 2(λ + 1)x + 3y + (1 + λ) = 0 ……………………. (4)
The centre of this circle is (-(λ + 1), \(\frac{3}{2}\)).
For the circle (4), 2x + 1 = 0 is one chord. This chord will be a diameter of the circle (4) if the centre of (4) lies on (3).
∴ 2{-(λ + 1)} + 1 = 0
⇒ λ = –\(\frac{1}{2}\)
Thus equation of the circle whose diameter is the common chord (1) and (2)
(put λ = \(\frac{1}{2}\) in equation (4))
2(x² + y²) + 2x + 6y + 1 = 0

Question 21.
Let us find the equation of a circle which cuts each of the following circles orthogonally
S’ ≡ x² + y² + 3x + 2y + 1 = 0 ………………… (1)
S” ≡ x² + y² – x + 6y + 5 = 0 …………….. (2)
and S” ≡ x² + y² + 5x – 8y + 15 = 0 ……………………. (3)
Solution:
The centre of the required circle is radical centre of (1), (2) and (3) and the radius is the length of the tangent from this point to any one of the given three circles. First we shall find the radical centre. For, the radical axis of (1) and (2) is
x – y = 1 ………………… (4)
and the radical axis of (2) and (3) is
3x – 7y = -5 …………………… (5)
The point of intersection (3, 2) of (4) and (5) is the radical centre of the circles (1), (2) and (3). The length of tangent from (3. 2) to the circle (1)
= \(\sqrt{3^{2}+2^{2}+3(3)+2(2)+1}=3 \sqrt{3}\)
Thus the required circle is
(x – 3)² + (y – 2)² = (3\(\sqrt{3}\))²
x² + y² – 6x – 4y – 14 = 0.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 10 యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు Formulas

→ యాదృచ్ఛిక చలరాశి : ఒక యాదృచ్ఛిక ప్రయోగం శాంపిల్ ఆవరణం S అనుకుందాం. ఏదైనా ప్రమేయం X : S → R ను యాదృచ్ఛిక చలరాశి అంటాం.

→ సంభావ్యతా విభాజన ప్రమేయం : X ఒక యాదృచ్ఛిక చలరాశి అప్పుడు F : R → R ప్రతి X ∈ Rకు F(x) = P(X ≤ x) తో నిర్వచితమైన ప్రమేయాన్ని X కు సంభావ్యతా విభాజన ప్రమేయం అంటాం.

→ X : S → R ఒక యాదృచ్ఛిక చలరాశి. X వ్యాప్తి పరిమితం లేదా అపరిమిత గణ్యసమితి అయితే X ను విచ్ఛిన్న చలరాశి అని, అట్లా కాకపోతే అవిచ్ఛిన్న యాదృచ్ఛిక చలరాశి అని అంటాం.

→ X : S → R ఒక విచ్ఛిన్న యాదృచ్ఛిక చలరాశి. దాని వ్యాప్తి = {X₁, X₂, ….} అయిన \(\sum_{r=1}^n P\left(X_r\right)\) = 1, P(X_r) ≥ 0.

→ X ఒక విచ్ఛిన్న యాదృచ్ఛిక చలరాశి. దాని వ్యాప్తి {X₁, X₂, ……} అనుకుందాం. ప్రతి n కు P(X = x_n) తెలిసి, Σx_n P(X = x_n) అనే మొత్తం పరిమితమైతే, ఆ మొత్తాన్ని X కు మధ్యమం (లేదా సగటు) అంటాం. దీన్ని µ తో సూచిస్తాం. (i.e.,) µ = Σx_n P(X = x_n)
Σ(x_n – µ)² P(X = x_n) అనేది ఒక పరిమిత సంఖ్య అయితే ఆ మొత్తాన్ని X కు విస్తృతి అంటాం.

→ X విస్తృతిని σ² తో సూచిస్తే, σ ను X కు క్రమ విచలనం అంటారు.
∴ μ = Σx_n P(X = x_n); σ² = Σ(x_n – μ)² P(X = x_n) = \(\Sigma x_n^2 P\left(X=x_n\right)\) – μ²

→ ద్విపద విభాజనం: n ఒక ధన పూర్ణాంకం. p వాస్తవ సంఖ్య మరియు 0 ≤ p ≤ 1. యాదృచ్ఛిక చలరాశి వ్యాప్తి {0, 1, 2, 3, …. n}. X ద్విపద చలరాశి లేదా ద్విపద విభాజనాన్ని పాటిస్తూ, n, p లు పరామితులుగా గల్గి వుంటే, P(X = r) = ⁿC_r . p^r q^n-r; r = 0, 1, 2, ….. n మరియు q = 1 – p అవుతుంది.

→ ద్విపద విభాజనాన్ని X ~ B(n, p) గా లేదా P(X = r) = ⁿC_r p^r q^n-r . p^r, r = 0, 1, 2, 3, …. n లేదా (q + p)ⁿతో సూచిస్తాం.

→ ద్విపద విభాజనం యొక్క మధ్యమం = np. ద్విపద విభాజనం యొక్క విస్తృతి = npq, క్రమవిచలనం = \(\sqrt{n p q}\).

→ పాయిజాన్ విభాజనం : λ > 0 ఒక స్థిరరాశి. యాదృచ్ఛిక చలరాశి X యొక్క వ్యాప్తి {0, 1, 2, ….}
P(X = k) = \(\frac{\lambda^k}{k !} e^{-\lambda}\), (k = 0, 1, 2, …….) అనుకుంటే, λ పరామితిగా, X పాయిజాన్ విభాజనాన్ని అనుసరిస్తుందని అంటాం. X ను పాయిజాన్ యాదృచ్ఛిక చలరాశి అంటాం. పాయిజాన్ విభాజనానికి మధ్యమము = λ, విస్తృతి = λ, క్రమ విచలనం = √λ

→ ఈ క్రింది షరతులకు లోబడి పాయిజాన్ విభాజనాన్ని, ద్విపద విభాజనపు సమతాస్థితి (limiting case) గా ఉజ్జాయింపు చేయవచ్చు.

• యత్నాల సంఖ్య n అనిశ్చితమైనంత పెద్దది, అంటే n → ∞
• ప్రతి యత్నంలో గెలుపు సంభావ్యత (స్థిరం) అతిస్వల్పం, అంటే p → 0

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 9 సంభావ్యత to solve questions creatively.

AP Intermediate 2nd Year Maths 2A సంభావ్యత Formulas

→ యాదృచ్ఛిక ప్రయోగం : ఒక ప్రయోగంలో ఏ ఫలితం వస్తుందో ముందే చెప్పలేనిదై, ఆ ప్రయోగ ఫలితాల జాబితా ముందే తెలిసి ఉండి, ఒకే విధమైన పరిస్థితుల్లో ఆ ప్రయోగాన్ని ఎన్నిసార్లైనా చేయడానికి వీలుంటే ఆ ప్రయోగాన్ని యాదృచ్ఛిక ప్రయోగం అంటాం.

→ లఘుఘటన, ఘటన ఒక యాదృచ్ఛిక ప్రయోగంలోని ఫలితాన్ని లఘుఘటన అంటాం. కొన్ని లఘు ఘటనల సమూహాన్ని ఘటన అంటాం.

→ పరస్పర వివర్జిత ఘటనలు: రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఏదైనా ఒక ఘటన సంభవించడం, మిగతా ఘటనల సంభవాన్ని నిరోధించేటట్టుంటే, అటువంటి ఘటనలను పరస్పర వివర్జిత ఘటనలు అంటారు.

→ సమ సంభవ ఘటనలు: రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఏ ఘటన అయినా మిగతా ఘటనల కంటే ఎక్కువగా సంభవిస్తుందనడానికి కారణమేమి లేకపోతే, అటువంటి ఘటనలను సమసంభవ ఘటనలు అంటారు.

→ పూర్ణ ఘటనలు : ఒక యాదృచ్ఛిక ప్రయోగంలోని రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఆ ప్రయోగం ఫలితం వాటిలో ఒక్కదానికైనా చెందేటట్లుంటే, అటువంటి ఘటనలను పూర్ణ ఘటనలు అంటాం.

→ సంభావ్యత సాంప్రదాయిక (లేదా గణితాత్మక) నిర్వచనం : ఒక యాదృచ్ఛిక ప్రయోగంలో n పూర్ణ, పరస్పర వివర్జిత, సమసంభవ ఘటనలుండి వాటిలో ఏదైనా ఘటన E జరగడానికి m అనుకూల ఫలితాలుంటే, ఆ ఘటన సంభావ్యతను P(E) తో సూచిస్తూ, P(E) = \(\frac{m}{n}\) గా నిర్వచిస్తాం. 0 ≤ P(E) ≤ 1

→ శాంపిల్ ఆవరణ : ఒక యాదృచ్ఛిక ప్రయోగపు ఫలితాన్ని లఘుఘటన అంటాం. ఒక యాదృచ్ఛిక ప్రయోగంలోని ఫలితాలన్నింటితో కూడిన సమితిని ఆ యాదృచ్ఛిక ప్రయోగానికి సంబంధించి శాంపిల్ ఆవరణ అంటాం. దీనిని S తో సూచిస్తాం. S లోని మూలకాలను శాంపిల్ బిందువులు అంటాం. S లో ప్రతిమూలకం, యాదృచ్ఛిక ప్రయోగం యొక్క ఒక ఫలితం S. ఒక ఉపసమితిని ఘటన అంటాం. అంటే, ఒక లఘుఘటనల సమితినే ఘటన అంటాం.

→ S శాంపిల్ ఆవరణ E ⊂ S.E లో ఒకే ఒక మూలకం ఉంటే E ని లఘుఘటన అంటాం. ఒక ప్రయోగ ఫలితం ఘటన E సంభవించింది లేదా జరిగింది అంటాం. అలాకాకపోతే ఘటన E సంభవించలేదు అంటాం.

→ Φ, S లు S కి ఉపసమితులు. వాటిని వరసగా అసంభవ ఘటన, నిశ్చిత ఘటన అంటాం.

→ ఘటన E కి పూరక ఘటనను E^C తో సూచిస్తాం. E^C = S – E శాంపిల్ ఆవరణ S కు E₁, E₂ లు రెండు ఘటనలు. అంటే E₁ ⊆ S, E₂ ⊆ S మరియు E₁ ∩ E₂ = Φ అయితే E₁, E₂ లను పరస్పర వివర్జిత ఘటనలు అంటాం.

→ ఘటనలు E₁, E₂, ……., E_n లు i ≠ j లకు 1 ≤ i, j ≤ n లకు E_i ∩ E_j = Φ అయ్యేటట్లుంటే, వాటిని పరస్పర వివర్జిత ఘటనలు అంటాం.

→ ఘటనలు E₁, E₂, …… E_k లు E₁ ∪ E₂ ∪ E₃ ∪ E_k = S అయితే వాటిని పూర్ణ ఘటనలు అంటాం. శాంపిల్ ఆవరణ S లో E₁, E₂లు రెండు ఘటనలు పూరక ఘటనలైన E₁ ∪ E₂ = S, E₁ ∩ E₂ = Φ.

→ ఒక యాదృచ్ఛిక ప్రయోగంకి శాంపిల్ ఆవరణ S. ఈ ప్రయోగపు అన్ని ఘటనల సమితి P(S) తో సూచిస్తాం. ఇచ్చట P(S), Sకు ఘాత సమితి.

→ సంభావ్యతా స్వీకృత నిర్వచనం: ఒక యాదృచ్ఛిక ప్రయోగపు శాంపిల్ ఆవరణ S. ప్రమేయం P : P(S) → R

→ క్రింది స్వీకృతాలను ధ్రువపరిస్తే Pని సంభావ్యతా ప్రమేయం అంటాం.

• P(E) ≥ 0 ∀ E ∈ P(S) (ధన స్వీకృతం)
• P(S) = 1 (పూరణ స్వీకృతం)
• E₁, E₂ ∈ P(S), E₁ ∩ E₂ = Φ అయితే P(E₁ ∪ E₂) = P(E₁) + P(E₂) (సమ్మేళును స్వీకృతం)
• ప్రతి E ∈ P(S) కు P(E) ను ఘటన E సంభావ్యత అంటాం.

→ శాంపిల్ ఆవరణ Sలో E ఒక ఘటన అయితే 0 ≤ P(E) ≤ 1. శాంపిల్ ఆవరణ Sలో E ఒక ఘటన అయితే

• P(E) : P(\(\bar{S}\)) ను E కు అనుకూలత అని
• P(E) : P(E) ను E కు ప్రతికూలత అంటాం.

→ సంభావ్యతపై సంకలన సిద్ధాంతం: ఒక యాదృచ్ఛిక ప్రయోగంలో E₁, E₂ లు రెండు ఘటనలైతే P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)

→ శాంపిల్ ఆవరణ Sకు E₁, E₂ లు రెండు ఘటనలు.
P(E₂ – E₁) = P(E₂) – P(E₁ ∩ E₂) మరియు P(E₁ – E₂) = P(E₁) – P(E₁ ∩ E₂)

→ E₁, E₂, E₃ లు శాంపిల్ ఆవరణ Sలో మూడు ఘటనలు అయిన P(E₁ ∪ E₂ ∪ E₃) = P(E₁) + P(E₂) + P(E₃) – P(E₁ ∩ E₂) – P(E₂ ∩ E₃) – P(E₃ ∩ E₁) + P(E₁ ∩ E₂ ∩ E₃)

→ ఒక యాదృచ్ఛిక ప్రయోగపు ఘటనలు A, Bలు అనుకుందాం. అప్పుడు “A జరిగిన తరువాత B జరగడం” అనే ఘటనను నియత ఘటన అంటాం. దీనిని \(\frac{B}{A}\) తో సూచిస్తాం. ఇట్లే \(\frac{A}{B}\) అనే ఘటన “B జరిగిన తరువాత A జరగడం” అనే ఘటనను సూచిస్తుంది.

→ నియత సంభావ్యత : ఘటన A జరిగిందని ఇస్తే, ఘటన B జరిగే సంభావ్యతను P(B/A) తో సూచిస్తాం. P(B/A) ని నియత సంఖ్య అంటాం.
దీనిని P(B/A) = \(\frac{P(B \cap A)}{P(A)}\), P(A) > 0 గా నిర్వచిస్తాం. ఇట్లే P(A/B) = \(\frac{P(B \cap A)}{P(B)}\), P(B) > 0.
సూచన : P(A/B) = \(\frac{n(A \cap B)}{n(B)}\); P(B/A) = \(\frac{n(B \cap A)}{n(A)}\)

→ నియత సంభావ్యతకు గణన సిద్ధాంతం ఒక శాంపిల్ ఆవరణం S లోని ఘటనలు A, Bలు ; P(A) > 0, P(B) > 0, అయితే P(A ∩ B) = P(A) . P(B/A) = P(B) . P(A/B).

→ స్వతంత్ర ఘటనలు : రెండు ఘటనలు A, Bలు P(A ∩ B) = P(A) . P(B) అయితే వాటిని స్వతంత్ర ఘటనలు అంటాం. అలాకాకపోతే A, B లను అస్వతంత్ర ఘటనలు అంటాం.

→ బేయీ సిద్ధాంతం: ఒక ప్రయోగంలో E₁, E₂, ……., E_n లు పరస్పర వివర్జిత, పూర్ణ ఘటనలవుతూ, P(E_i) > 0, i = 1, 2, …. n అయినప్పుడు k = 1, 2, 3, ….. n లకు
\(P\left(\frac{E_k}{A}\right)=\frac{P\left(E_k\right) \cdot P\left(\frac{A}{E_k}\right)}{\sum_{i=1}^n P\left(E_i\right) \cdot P\left(\frac{A}{E_i}\right)}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 8 విస్తరణ కొలతలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A విస్తరణ కొలతలు Formulas

→ అవర్గీకృత దత్తాంశానికి మధ్యమం = \(\frac{విచలనాల మొత్తం}{పరిశీలనల సంఖ్య}\) = \(\frac{\Sigma x_i}{n}\)

→ అవర్గీకృత దత్తాంశానికి మధ్యగతం: ముందుగా దత్త n పరిశీలనలను పరిమాణపరంగా అవరోహణ లేదా ఆరోహణ క్రమంలో వ్రాయవలెను.

→ n బేసి సంఖ్య అయితే \(\frac{n+1}{2}\) వ పరిశీలనల అవర్గీకృత దత్తాంశం యొక్క మధ్యగతం అగును.

→ n సరి సంఖ్య అయితే \(\frac{n}{2}\) మరియు \(\frac{n+2}{2}\) వ పరిశీలనల సరాసరిను అవర్గీకృత దత్తాంశం యొక్క మధ్యగతం అగును.

→ వర్గీకృత మరియు అవర్గీకృత దత్తాంశానికి వ్యాప్తి, మధ్యమ విచలనం, విస్తృతి మరియు ప్రామాణిక విచలనం కొన్ని విస్తరణ కొలతలు

→ వ్యాప్తిని దత్తాంశ గరిష్ట విలువకు, కనిష్ఠ విలువకు మధ్యగల భేదంగా నిర్వచిస్తారు.

→ అవర్గీకృత విభాజనానికి మధ్యమ విచలనం

• మధ్యమం నుంచి మధ్యమ విచలనం = \(\frac{\sum\left|x_i-\bar{x}\right|}{n}, \bar{x}\) మధ్యమం
• మధ్యగతం నుంచి మధ్యమ విచలనం = \(\frac{\sum \mid x_i-\text { మధ్యగతం } \mid}{n}\)

→ వర్గీకృత దత్తాంశానికి మధ్యమ విచలనం

• మధ్యమం నుంచి మధ్యమ విచలనం = \(\frac{1}{N} \sum f_i\left|x-\bar{x}_i\right|\); N = Σf_i, మరియు \(\bar{x}\) మధ్యమం
• మధ్యగతం నుంచి మధ్యమ విచలనం = \(\frac{1}{N} \sum f_i \mid x_i\) – మధ్యగతం|, N = Σf_i

→ అవర్గీకృత దత్తాంశానికి విస్తృతి, σ² = \(\frac{1}{n}\) = \(\sum\left(x_i-\bar{x}\right)^2\), ప్రామాణిక విచలనం σ = \(\sqrt{\frac{1}{n} \sum\left(x_1-\bar{x}\right)^2}\)

→ విచ్ఛిన్న పౌనఃపున్య విభాజనానికి విస్తృతి, σ² = \(\frac{1}{N}\) = \(\sum f_i\left(x_i-\bar{x}\right)^2\), \(\bar{x}\) మధ్యమం

→ ప్రామాణిక విచలనం σ = \(\sqrt{\frac{1}{N} \sum f_i\left(x_i-\bar{x}\right)^2}\)

→ అవిచ్ఛిన్న పౌనఃపున్య విభాజనానికి ప్రామాణిక విచలనం σ = \(\frac{1}{N} \sqrt{N \sum f_i x_i^2-\left(\sum f_i x_i\right)^2}\) (లేదా) σ = \(\frac{h}{N} \sqrt{N \Sigma f_i y_i^2-\left(\Sigma f_i y_i\right)^2}, y_i=\frac{x_i-A}{h}\)

→ విచలనాంకం = \(\frac{\sigma}{x} \times 100(\bar{x} \neq 0)\)

→ దత్తాంశంలోని ప్రతి పరిశీలనను ఒక స్థిరరాశి K చే గుణించినపుడు ఫలితంగా వచ్చే పరిశీలనల విస్తృతి, తొలిపరిశీలనల విస్తృతికి K² రెట్లు ఉంటుంది.

→ పరిశీలనలు x₁, x₂, …….., x_n లలో ప్రతిదానిని K కి పెంచితే లేదా కలిపితే (K ఒక ధనాత్మక లేదా ఋణాత్మక సంఖ్య), వచ్చే పరిశీలనల విస్తృతి మారదు.