topic: class +2 - dr. sangeeta khannasolution dr. sangeeta khanna ph.d 1 chemistry coaching circle...

$: TOPIC: CLASS +2 - Dr. Sangeeta KhannaSOLUTION Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE G:\Obj-Solution.doc TOPIC: CLASS +2 INSTRUCTIONS 1. The test is of 1 hour 30 minutes$
Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE G:\Obj-Solution.doc

TOPIC: SOLUTION CLASS +2

INSTRUCTIONS

1. The test is of 1 hour 30 minutes duration.

2. The maximum marks are 160.

3. This test consists of 50 questions.

4. For each question in Section A, B & C you will be awarded 3 marks if you have darkened only the

bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (-1) mark will be awarded.

5. For each question in Section D you will be awarded 3 marks if you darken the bubble corresponding to

the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for in

this Section.

6. For each question in Section E, you will be awarded 2 marks for each row in which you have darkened

the bubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s) in this Section.

7. For each question in Section F, you will be awarded 3 marks if you darken the bubble corresponding

to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for

in this Section.

SECTION – A (Single Correct Choice Type)

This Section contains 23 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 23 × 3 = 69 Marks

1. The solubility of a specific non-volatile salt is 4g in 100 g of water at 250C. If 2.0 g , 4.0 g and 6.0 g of the

salt added to 100 g of water at 250C, in system X, Y and Z. The vapour pressure would be in the order: a. X < Y < Z b. X > Y > Z c. Z > X = Y d. X > Y = Z D Sol. Solution X is unsaturated so vapour pressure will be more, solution Y & Z are saturated so vapour pressure

of Y = vapour pressure of Z and 2 gm of solute would be present in form of solid in system Z. 2. A living cell contains a solution which is isotonic with 0.2 M glucose solution. What osmotic pressure

develops when the cell is placed in 0.05 M BaCl2 solution at 300 K? a. 1.23 atm b. 3.69 atm c. 6.15 atm d. 2.46 atm A Sol. = (iC1 – i2C2)RT (1 × 0.2 – 3 × 0.05) 0.0821 × 300 1.23 atm 3. If potassium iodide, KI, behaves as though it were completely dissociated in solution, the freezing point

at a one molal aqueous solution be (-3.72°C). (The depression should be twice of Kf because one mole of KI provides two moles of ions in solution.) Mercuric iodide, Hgl2, is now dissolved in the solution until

all the iodide ion has reacted according to HgI2 + 2I– [HgI4]

-2. What should be the freezing point of the new solution.

a. It will be increased due to decrease in number of particle. b. It will be decreased due to decrease in number of particle after dissolution. c. It will be increased due to increase in number of particle after dissolution. d. None A 4. Van’t Hoff factor of 0.01 M CH3COOH solution is 1.04. What will be the ionisation constant and pH of

solution Ka pH Ka Ph Ka pH Ka pH a. 1.27× 10-5 2.9 b.1.57 × 103 5.2 c. 1.6 × 10-5 3.4 d.1.6×10-6 4.10 C Sol. CH3COOH CH3COO + H+ i = 1 + = 0.04

Ka = C2 i.e. = 0.01 × (0.04)2 = 16 × 10-6

H+ = C = 0.01 × 0,04 = 4 × 10–4

Dr. Sangeeta Khanna Ph.D


5. Two solutions labeled as 3M HCl and 1M HCl are mixed in the ratio of x : y by volume and the molarity of

mixture solution becomes 1.5 M. What is the molarity of the resulting solution if they are mixed in the ratio of y : x by volume?

a. 4 M b. 3M c. 2M d. 2.5M D

Sol. 1.5 = yx

yx3

or 1.5x + 1.5y = 3x + y or 1.5x = 0.5y or

3

1

y

x

when the solution is mixed in the ratio x

y i.e.,

1

3

M = 4

10

4

1133

= 2.5 M

6. Benzene (p° = 160 mm) and toluene (p° = 68 mm) form ideal solution at certain temperature with mole fraction of benzene as 0.2. The vapour pressure of solution will be

a. 220 mm b. 86.4 mm c. 160 mm d. cannot be predicted B

Sol. pBenzene = 160 × 0.2 = 32 mm, pToluene = 68 × 0.8 = 54.4 mm p = 32 + 54.4 = 86.4 mm 7. Which of the following have equal van’t Hoff factor?

a. 0.01 M K3[Fe(CN)6] and 0.10 M FeSO4 b. 0.10 M K4[Fe(CN)6] and 0.10 M FeSO4. (NH4)2SO4. 6H2O

c. 0.30 M NaCl and 0.20 M BaCl2 d. 0.05 M K2SO4. Al2(SO4)3. 24H2O and 0.02 M KCl. MgCl2 . 6H2O

B

Sol. 4664 ])CN(Fe[K4])CN(Fe[K (i=5)

FeSO4. (NH4)2SO4. 6H2O Fe2+ + SO42– + 2NH4+ + SO4

2–+ 6H2O (i = 5) Both K4[Fe(CN)6] and FeSO4. (NH4)2SO4. 6H2O gives same number of ions on complete ionization hence they have same van’t Hoff factor.

8. 0.1 molal aqueous solution of an electrolyte AB3 is 90% ionised. The boiling point of the solution at 1 atm is:

[Kb(H2O) = 0.52 K Kg mol–1]

a. 273.19 K b. 374.92 K c. 376.4 K d. 373.19 K D

Sol. i = 1 + 3 1 × 3 × 0.9 3.7;

Tb = i Kb.m = 0.52 × 0.1 × 3.7 = 0.19

TB = 0b

T + 0.19 = 373 + 0.19 = 373.19 K

9. A solution of LiCl in water has XLiCl = 0.0800. What is the Molality? a. 4.01 m LiCl b. 4.44 m LiCl c. 4.83 m LiCl d. 8.70 m LiCl C 10. Identify the incorrect statement. (i) Ebullioscopic constant is independent of the solvent. (ii) Isotonic solutions always have same osmotic pressure and same molar concentration at a given

temperature for electrolytic solution.

(iii) Higher freezing point depression of 0.01 M NaC than 0.01M AlC3 is owed to smaller value of van’t

Hoff factor for NaC.

a. (i) b. (ii) c. (iii) d. All are incorrect D 11. The relationship between the values of osmotic pressure of 0.1 M solution oxalic acid (P1) and acetic

acid (P2) is [K1 (oxalic acid >> K2 (oxalic acid) and K1 = 10-2, Kacetic acid = 10–5 ].

a. P1 > P2 b. P1 = P2 c. P2 > P1 d. 2

21

22

221

21

)PP(

P

)PP(

P



A 12. The dichromate ion oxidses stannous ions in solutions according to the reaction

3Sn2+ + 14H+ + Cr2O 27

3Sn4+ + 2Cr3+ + 7H2O

How many mL of K2Cr2O7 solution of 0.5 M strength will be required to completely react with 0.5 mol of

tin (II) chloride in solution

a. 666.6 mL b. 444.4 mL c. 222.2 mL d. 333.3 mL

D

Sol. From the given equation

3 mol of Sn2+ requires K2Cr2O7 = 1 mol

0.5 mol of Sn2+ requires K2Cr2O7 = 3

5.0mol

Now, 5.010V3

5.0 3)mL( or V = 333.3 mL

13. At 17°C, the osmotic pressure of sugar solution is 580 torr. The solution is diluted and the temperature is

raised to 57°C, when the osmotic pressure is found to be 165 torr. The extent of dilution is:

a. 2 times b. 3 times c. 4 times d. 5 times

C

Sol. 1 = 1

1

V

nRT, 2 =

2

2

V

nRT

Hence, 330V

290V

TV

TV

165

580

1

2

21

12

2

1

1

2

V

V= 4

14. 1000 g of a sample of hard water was found to contain 0.01 g of MgSO4. The concentration of MgSO4 is

a. 100 ppm b. 10 ppm c. 1 ppm d. 103 ppm

B

Sol. Concentration of Mg2+ = 6101000

01.0 = 10 ppm.

15. Solubility of oxygen gas in water follows Henry’s law. When

the solubility is plotted against partial pressure at a definite

temperature we get following plots.

Which of the following sequence of temperatures is correct?

a. T1 = T2 = T3 = T4

b. T1 > T2 > T3 > T4 c. T1 < T2 < T3 < T4 d. T1 > T2 < T3 > T4

B 16. A solution is prepared containing a 2 : 1 mol reatio of dibromo ethane (C2H4Br2) and dibromo propane

(C3H6Br2) what is the total vapour pressure over the solution assuming ideal behaviour? Vapour pressure (mm Hg) C2H4Br2 173 C3H6Br2 127

a. 300 mm Hg b. 158 mm Hg c. 150 mm Hg d. 142 mm Hg B Sol. C2H4Br2(A); C3H6Br2(B);

3

2XA

3

1XB

0A

P 173 mm 0B

P = 127 mm

P = PA + PB

= B0BA

0A XPXP

= 173 × 3

2 + 127 ×

3

127346

3

1

= 3

473 = 157.6 158 mm

Partial Pressure of O2

Solubility of O2 in water T1

T2

T3

T4



17. Insulin (C6H10O5)n dissolved in a medium shows osmotic pressure (atm) at C gm/lit concentration and

300 K temperature. The slope of plot of against C is 4.1 × 10–3. Molecular mass of insulin is:

a. 3 × 103 b. 6 × 106 c. 3 × 106 d. 6 × 103

B

Sol. nRT1000

V

RTm

1000wV

B

B

B

B

m

RT 1000

V

w

= C × Bm

RT 1000

When is plotted against C, the slope will be equal to Bm

RT 1000

4.1 × 10–3 = Bm

300082.01000

mB = 6 × 106

18. A solution weighing a g has molality b. The molecular mass of solute if the mass of solute is c g, will be:

a. )ca(

1000

b

c

b.

)ba(

1000

a

b

c.

)ca(

1000

c

b

d.

)ab(

1000

b

c

A

Sol. Molality, m = AB

B

w

1000

m

w

b = )ca(

1000

m

c

B

mB = )ca(

1000

b

c

19. Compound of PdCl4.6H2O is a hydrated complex; 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex: (Kf for water = 1.86 K kg mol–1)

a. [Pd(H2O)6]Cl4 b. [Pd(H2O)4Cl2]Cl22H2O

c. [Pd(H2O)3 Cl3]Cl3H2O d. [Pd(H2O)2Cl4]4H2O C

Sol. T = i × Kf × m (273 – 269.28) = i × 1.86 × 1 3.72 = i × 1.86 i = 2

It give two ions in solution like in (C)

Thus, the complex should give two ions in the solution, i.e., the complex will be [Pd(H2O)3Cl3]Cl3H2O.

20. Total vapour pressure of mixture of 1 mol A ( 0A

P = 150 torr) and 2 mol B( 0B

P = 240 torr) is 200 torr. In this

case:

a. there is positive deviation from Raoult’s law b. there is negative deviation from Raoult’s law c. there is no deviation from Raoult’s law d. molecular masses of A and B are also required for calculating the deviation B

Sol. XA = 3

1, XB =

3

2

P = B0BA

0A XPXP

= mm 210160503

2240

3

1150

Pexp. < Pcalculated

There is negative deviation from Raoult’s law.



21. Which statement about the composition of vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume the temperature is constant at 25°C. Vapour pressure data (25°C):

Benzene 75 mm Hg Toluene 22 mm Hg a. The vapour will contain higher percentage of benzene

b. The vapour will contain higher percentage of toluene c. The vapour will contain equal amount of benzene and toluene d. Not enough information is given to make a prediction A

Sol. A : Benzene B : Toluene P = PA + PB

P = B0BA

0A XP XP

= 2

122

2

175

= 37.5 + 11 = 48.5

Mole fraction of benzene in vapour, YA = 48

5.37

P

PA = 0.78

Similarly, mole fraction of toluene in vapour, YB = 0.22

The vapour will contain higher percentage of benzene. 22. The vapour pressure of water at T (K) is 20 mm Hg. The following solutions are prepared at T (K):

I. 6 g of urea (mol. wt. = 60) is dissolved in 178.2 g of water. II. 0.01 mol of glucose is dissolved in 179.82 g of water III. 5.3 g of Na2CO3 (mol. wt. = 106) is dissolved in 179.1 g of water. Identifying the correct order in which the vapour pressures of solutions increases: a. III, I, II b. II, III, I c. I, II, III d. I, III, II A

Sol. I. XB = 01.0

60

6

18

2.17860

6

nn

n

BA

B

01.0XP

PB

0

II. 001.0

01.018

82.179

01.0

nn

nX

BA

BB

001.0P

P

0

III. 005.0

106

3.5

18

1.179106

3.5

nn

nX

BA

BB

B0

X iP

P

= 3 × 0.005 = 0.015

Vapour pressure of solutions will increase in the following sequence. (III) < (I) < (II) 23. Which of the following is correct order of elevation in Boiling point.

(I) 1 N NaCl (II) 1 N Na2SO4 (III) 1 N Na3PO4 (IV) 1 N urea a. I < II < III < IV b. IV < I < III < II c. IV < III < II < I d. IV = III < I < II C

Sol. Convert normality to molarity & then calculate (i × M):

(i) 1 × 2 = 2 (ii) 5.132

1 (iii) 33.14

3

1 (vi) 1



SECTION – B (Assertion-Reason Type)

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. 5 × 3 = 15 Marks

(a) Statement-1 and Statement -2 are true, Statement-2 is a correct explanation of Statement-1. (b) Statement-1 and Statement -2 are true, Statement-2 is not a correct explanation of Statement-1. (c) Statement-1 is true and Statement-2 is false. (d) Statement-1 is false and Statement-2 is true. 1. Statement-1: Camphor is used as solvent in the experimental determination of molecular masses of

napthalene and anthracene Statement-2: Camphor has high molal elevation constant a. (a) b. (b) c. (c) d. (d) C

Sol. Camphor does not have high molal elevation constant. 2. Statement-1: The vapour pressure of 0.5 molal urea and 0.5 molal NaCl solution will be different.

Statement-2: The vapour pressure of solution having non-volatile solute may be calculated as: P = P0XA P0 = Vapour pressure of pure solvent, XA = Mole fraction of solvent. a. (a) b. (b) c. (c) d. (d) B

3. Statement -1: Desalination is the process of reverse osmosis, used to get the drinking water from sea-water. Statement-2: When sea-water is separated from normal water using semipermeable membrane and high pressure is applied to the side of sea-water then water molecules pass towards normal water. a. (a) b. (b) c. (c) d. (d) A

Sol. Desalination is the process of reverse osmosis. On applying extra pressure to solution side, the osmosis takes place in reverse direction.

4. Statement-1: Observed colligative property of acetic acid dissolved in benzene is found greater than calculated colligative property. Statement-2: Acetic acid undergoes association to form dimmer, when dissolved in benzene a. (a) b. (b) c. (c) d. (d) D

Sol. When acetic acid is dissolved in benzene, it undergoes dimerisation thus colligative properties must decrease. 5. Statement-1: Rate of evaporation of water is less than that of ether under identical conditions.

Statement-2: Stronger is the intermolecular force, slower is the rate of evaporation and H2O has stronger interparticle forces. a. (a) b. (b) c. (c) d. (d) A

SECTION – C (Paragraph Type)

This Section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the second paragraph 4 multiple choice questions have to be answered. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 6 × 3 = 18 Marks

Passage-1 A solution is prepared by mixing ‘A’ & ‘B’. Its B.pt. mole fraction graphs is given as under

TA

XA = 1 X4 = .4

XB = .6

XB = 1

B A

C

TB



1. Which of the following statement is incorrect: a. A composition with 50 mole % of A component will give vapours rich with A & remaining liquid rich

with Azeotrope. b. A composition with C on evaporation will give vapours of azeotrope. c. A composition in between A & C on distillation will give vapours richer with B until azeotropic mixture

is reached. d. A mixture of A & B with mole % 40 & 60 respectively have same composition in vapour phase. A 2. Pure B can be obtained by distillation of a. Any molar fraction between A & B b. Molar fraction between A & C c. Molar fraction between B & C d. Molar fraction with 40% A & 60% B.

C Passage – 2 The boiling point elevation and the freezing point depression of solution have a number of practical

applications. Ethylene glycol (CH2OHCH2OH) is used in automobile radiators as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as antifreeze. For boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol (CH3OH), a fairly volatile liquid that boils only at 65°C is sometimes used as antifreeze in automobile radiators.

3. Which of the following is a better reagent for depression in freezing point but not for elevation in boililng

point?

a. CH3OH b. OHCH

OHCH|

2

2

c. HOH

OHCH

OHCH

C

2

2

|

| d. C6H12O6

A 4. 124 g each of the two reagents glycol and glycerol are added in 5 kg water of the radiators in the two

cars. Which of the following statements is wrong?

a. Both will act as antifreeze b. Glycol will be better c. Glycerol is better because its molar mass is greater than glycol d. Glycol is more volatile than glycerol C

5. 620 g glycol is added to 4 kg water in the radiator of a car. What amount of ice will separate out at -6°C?

Kf = 1.86 K Kg mol–1:

a. 800 g b. 900 g c. 600 g d. 1000 g

B

Sol. T = Kf × AB

B

wm

1000w

6 = 1.86 × Aw62

1000620

WA = 3100 g (Mass of water)

Amount of ice = 4000 – 3100 = 900 g

6. If cost of glycerol, glycol and methanol per kg are same, then the sequence of economy to use these

compounds as antifreeze will be:

a. glycerol > glycol > methanol b. methanol > glycol > glycerol

c. methanol = glycol = glycerol d. methanol > glycol < glycerol

B



SECTION – D (Multiple Correct Choice Type)

This Section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. 8 × 3 = 24 Marks

1. Consider 0.1 M solutions of two solutes X and Y. The solute X behaves as univalent electrolyte, while the

solute Y dimerises in solution. Select correct statement(s) regarding these solutions: a. The boiling point of solution of ‘X’ will be higher than of ‘Y’ b. The osmotic pressure of solution of ‘Y’ will be lower than that of ‘X’ c. The freezing point of solution of ‘X’ will be lower than that of ‘Y’ d. The relative lowering of vapour pressure of both the solution with be the same A, B, C 2. Which one of the following statement is True? a. The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is BaCl2 >

KCl > CH3COOH > sucrose

b. The osmotic pressure () of a solution is given by the equation = MRT, where M is the molarity of the solution

c. Raoult’s law states that the vapour pressure of a volatile component over a solution is proportional to its mole fraction in solution

d. Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression.

A, B, C

Sol. Tf = I kf m Since Kf is different in different solvents, Tf will be different. 3. Study the figure given aside and pick out the correct option(s) of the following

a. A white precipitate of PbSO4 is formed on Na2 SO4 side b. A white precipitate of PbSO4 is formed on PbCl2 side c. No precipitate is formed on either side d. Meniscus of PbCl2 solution rises and that of Na2 SO4 solution

falls in due course of time C, D 4. Dry air is slowly passed through three solution of different concentrations, c1, c2 and c3 ; each containing

(non volatile) NaCl as solute and water as solvent, as shown in the Fig. If the vessel 2 gains weight and the vessel 3 loses weight, then

a. c1 > c2 b. c1 < c2 c. c1 < c3 d. c2 > c3 B, D

5. Which of the following statements is/are correct about azeotropic mixture?

a. Azeotropic mixtures are non-ideal solution b. All the components of azeotropic mixture cannot be separated by fractional distillation c. Azeotropes obey Raoult’s law d. Solution with positive deviation from Raoult’s law, forms minimum boiling azeotrope A, B, D

6. Which of the following solutions exhibit positive deviation from Raoult’s law?

a. H2O + C2H5OH b. C6H6 + C2H5OH c. H2O + HCl d. CHCl3 + (CH3)2CO A, B

Sol. (H2O + C2H5OH) and (C6H6 + C2H5O) show positive deviation from Raoult’s law while (H2O + HCl) and [CHCl3 + )(CH3)2CO] show negative deviation from Raoult’s law.

0.1M Na2 SO4

0.2M PbCl2

Semipermeable membrane

Vessel - 1 Vessel - 2 Vessel - 3



7. Consider the following solutions: I. 1 M aqueous glucose II. 1 M aqueous NaCl III. 1 M C6H5COOH in C6H6 IV. 1 M (NH4)3PO4 a. All are isotonic solutions b. III is hypotonic to I, II, IV c. I, II, IV are hypertonic to III d. IV is hypertonic to I, II, III B,C,D

Sol. I, II and IV are hypertonic to III because benzoic acid undergoes association when dissolved in benzene. 8. Which of the following are correct about the binary homogeneous liquid mixture?

a. H2O and C2H5OH b. C6H6 and C6H5CH3

Hsol. > 0, Vsol. > 0 Hsol. = 0, Vsol. = 0 c. CH3COCH3 and CHCl3 d. H2O and HCl

Hsol. < 0, Vsol. < 0 Hsol. > 0, Vsol. < 0 A,B,C

Sol. (H2O + C2H5OH) Shows positive deviation from Raoult’s law. Therefore, Hmix > 0, Vmix > 0

(C6H6 + C6H5CH3) It is ideal solution therefore, Hmix = 0, Vmix = 0

(CH3COCH3 + CHCl3) Shows negative deviation from Raoult’s law, therefore, Hsol. < 0, Vsol. < 0 9. Three solutions of HCl having normality 12N, 6N and 2N are mixed to obtain a solution of 4N normality.

Which among the following volume ratio is correct for the above three components?

a. 1 : 1 : 5 b. 1 : 2 : 6 c. 2 : 1 : 9 d. 1 : 2 : 4 A,B,C

Sol. N1V1 + N2V2 + N3V3 = NR(V1 + V2 + V3) 12 × 1 + 6 × 2 + 2 × 6 = NR × 9 NR = 4

SECTION – E (Matrix Type)

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS. 2 × 8 = 16 Marks

1. Match Column – I with Column – II

Column – I Column – II

A. B. C. D.

1 : 0.1 M glucose; 2 : 0.1 M urea

1 : 0.1 M NaCl; 2 : 0.1 M Na2SO4

1 : 0.1 M NaCl; 2 : 0.1 M KCl

1: 0.1 M CuSO4; 2 : 0.1 M sucrose

p. q. r. s.

1 and 2 are isotonic. No net migration of solvent across the membrane

1 is hypertonic to 2

1 is hypotonic to 2

Sol. A p, q; B s; C p, q; D r Diagram based match (Q. 2)

B

L

A

L

V V

C E F

TB

TA

XA = 1 XB = 1

acetone Chloroform

2C6H5COOH Benzene

C6H5 C

O H O

O H O

C – C6H5

– +

+ – Dimer



2. Match the entries of column I with appropriate entries of column II. (Multiple Match)

Column – I Column – II

(A) (B) (C) (D)

At Composition C At composition E At composition F At composition between F & C

(p) (q) (r) (s) (t)

a mixture with positive deviation maximum Hydrogen bonding will be observed between solute & solvent a mixture with negative deviation Vapour are rich with CHCl3 & flask is rich with azeotrope. Vapour are rich with Acetone & flask is rich with azeotrope.

Sol. A q, r; B r, s; C r, t; D r, t

SECTION – F (Integer Type) This Section contains 5 questions. The answer to each question is a single digit integer ranging from 0

to 9. The correct digit below the question number in the ORS is to be bubbled. 5 × 3 = 15

1. Relative decrease in vapour pressure of an aqueous solution containing 2 mol NaCl in 3 mol H2O is 0.6.

On reaction with AgNO3, this solution will form ppt. of AgCl. How many moles of AgCl will be formed.

Sol. Calculated value of 4.032

2x

p

pB

Observed value of 6.0p

p

Van’t Hoff factor = 5.14.0

6.0

1

15.1

12

1i

= 0.5 or 50%

Since NaCl is 50% ionized

No. of mol of Cl– = 100

502 = 1

Mol of AgCl formed = 1 mol.

2. A 1.2% (W/V) solution of NaCl is isotonic with 7.2% (W/V)solution of glucose. The van’t Hoff factor of NaCl is nearly equal to:

Sol. RTiCNaCl 1

RTC2)ecosglu(

)ecosglu(NaCl

iC1RT = C2RT iC1 = C2

V180

2.7

V5.58

2.1i

95.11802.1

5.582.7i

2

3. Drawing (1) shows a system in which an equilibrium exists between dissolved and undissolved gas particles at P = 1 atm. According to Henery’s law, if the pressure is increased to 2 atm and equilibrium is restored, which drawing (2) – (5) best represents the equilibrium at 2 atm?

Sol. 5 4. K2[HgI4] is 50% ionized in aqueous solution. What will be its von’t Hoff’s factor.

Sol. i = 1 + 2

= 1 + 2 - 5 = 2 5. An aqueous solution containing ionic salt having molality equal to 0.1892 freezes at -0.704°C. The van’t

Hoff factor of the ionic salt is given by _______. (Kf = 1.86 Km–1) Sol. 2

1 atm

(1) (2)

2 atm

2 atm

(3)

2 atm

(5)

2 atm

(4)

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