Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh - Dr

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2017\+2\Physical\Grand Test\+2 Grand Test - 2 13.5.2017\Grand Test-2 Solution 13.5.2017.doc

Test Venue:

Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh

Dr. Sangeeta Khanna Ph.D


Grand Test – 2 (Solutions) Test – 3 9.5.2017 [Solutions]

READ INSTRUCTIONS CAREFULLY 1. The test is of 2 hour duration. 2. The maximum marks are 271. 3. This test consists of 65 questions. 4. Keep Your mobiles switched off during Test in the Halls.

SECTION – A (Single Correct Choice Type) Negative Marking [-1]

This Section contains 37 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 37 × 4 = 148 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

Sol. 2KI + HgI2 K2[HgI4] 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 CH3CO O + 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



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 Sol. Oxalic acid will dissociate more than acetic acid due to high K1 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 ratio 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

Partial Pressure of O2

Solubility of O2 in water T1

T2

T3

T4



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

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 in gm is:

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

Sol. = CRT; RTV

n ;

V

RT

M

W

B

B

= B

B

M

RT

V

W

= atm WB = gm V = lit. MB = M.Wt.

B

B

m

RT

V

w

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

RT

4.1 × 10–3 = Bm

300082.0

mB = 6 × 103 18. A solution weighing a gm has molality b. The molecular mass of solute if the mass of solute is c gm,

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

C (gm/lit.)



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. Benzene is more volatile then toulene 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 180 g of water

III. 5.3 g of Na2CO3 (mol. wt. = 106) is dissolved in 180 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. 001.001.10

01.0

01.010

01.0

01.018

180

01.0

nn

nX

BA

BB

001.0P

P

0

III. 0049.005.10

05.0

05.010

05.0

106

3.5

18

180106

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):

(ii) 2

NM ; (iii)

3

NM ; (iv) & (i) M = N

(i) 1 × 2 = 2 (ii) 5.132

1 (iii) 33.14

3

1 (vi) 1

24. The concentration of glucose solution in molality which boils at 100.01 °C (Kb for water is 0.5 °C kg mol–) is:

a. 0.01 b. 0.02 c. 0.03 d. 0.04 B Sol. We known that

Tb = Kb × m (molality of solution)

Here, Tb = 100.01 – 100 = 0.01 °C ; Kb = 0.5 °C kg mol–

0.01 °C = 0.5 °C kg mol– × m; m = 05.0

01.0 = 0.02 mol kg–

25. The elevation in boiling point for 13. 44 gm of CuCl2 dissolved in 1 kg of water as solvent will be: (Kb = 0.52 K kg mol ; molar mass of CuCl2 = 134.4 g/mol) a. 0.05 b. 0.10 c. 0.16 d. 0.20 C

Sol.

on)dissociati of degree ( 2

00

2

1

1

2

ondissociati after Conc.

(mol)Conc Initial

2Cl Cu CuCl .actionRe



Vant Hoff factor, i = 1 – + + 2 = 1 + 2

But elevation in b.pt, solvent of w t. solute of w t..mol

solute of w t.K1000iT b

b

Wt. of solvent = 1 kg 1 × 1000 g = 1000 g; i = 3

16.0156.010004.134

44.1352.010003T

b

26. Suppose the radiator of an automobile contains 11.0 L water. What is the freezing point on addition of 4.6 kg of prestone [glycol, C2H4(OH)2]? Also calculate the amount of zerone [methyl alcohol, CH3OH], that would be needed to produce the same result. Assume 100% purity of compounds added (Kf = 1.86 °C/m).

a. 3.37 kg; 12.5°C b. 2.37 kg; –12.5°C c. 2.37 gm; –12.5°C d. 23.7 gm; –10.4°C B

Sol. Mol. wt. of glycol C2H4(OH)2 = (2 × 12) + (4 × 1) + 2 (16 + 1) = 62 g mol–. Mol. wt. of methyl alcohol, CH3OH = 12 + (3 × 1) + 16 + 1 = 32 g mol–; wt. of glycol, W2 = 4.6 kg; Wt. of solvent, W1 = Volume ×

density = 11.0 L × 1 g (mL)– = 11 × 1000 mL × 1 g (mL)– = 11000 g; Kf = m

C86.1 = 1.86°C kg mol–;

Tf = ? we know that:

Tf =

kg1000

11000mol g 62

g 10006.4mol kg C 86.1

kg1000

g in Wmol g in M

g) (in W)mol kg C in(K

-

-

1-2

2-

f

Tf = 12.5 °C. Hence freezing point = -12.5 °C In order to get the same freezing point depression with CH3OH, we require same number of mol as of

glycol. But:

no. of mol of glycol = -- mol g 62

g 10006.4

mol g in glycerol of w t. mol

g) (in glycol of w t.

= 74.19 mol

no. of mol of CH3OH = 74.19 mol

wt. of CH3OH = 74.19 mol × g 1000

kg 1

mol 1

g 32 = 2.37 kg Ans.

27. Calculate the freezing point of a 10% (by weight) solution of ethyl alcohol (C2H5OH) in water. (Kf for water = 1.86 °C/m; at wt. C = 12, H = 1, O = 16).

a. + 4.49 °C b. - 4.49 °C c. 5.49 °C d. - 5.49 °C B

Sol. Wt. of C2H5OH, W2 = 10 g; wt. of water solvent, W1 = 100 – 10 = 90 g. Mol. wt. of C2H5OH = (2 × 12) + (5 × 1) + 16 + 1 = 46.0 g mol–, Kf = 1.86 °C/m = 1.86 °C kg mol–. We know that:

Tf = C 49.4

kg1000

90mol g 46

g 10mol kg C 86.1

kg1000

g in Wmol) g (in M

g) in(W)mol kg C in(K

-

-

1-2

2-

f

Freezing point = - 4.49°C Ans. 28. Solution A contains 7 g/L of MgCl2 and solution B contains 7 g/L of NaCl. At room temperature, the

osmotic pressure of

a. Solution A is greater than B b. Both have same osmotic pressure c. Solution B is greater than A d. Cannot determine C

Sol. MgCl2 Mg2+ + 2Cl– i.e., 1 + 2 = 3 particles NaCl Na+ + Cl– i.e., 1 + 1 = 2 particles



Osmotic pressure is a colligative property which depends upon the number of particles. The number of particles in MgCl2

being more than that in NaCl, MgCl2 solution will have higher osmotic pressure. 29. At some temperature, which pair of the following solutions are isotonic solutions?

a. 0.2 M BaCl2 and 0.2 M urea b. 0.1 M urea and 0.1 M NaCl c. 0.1 M NaCl and 0.1 M K2SO4 d. 0.1 M Ba(NO3)2 and 0.1 M Na2SO4 D

Sol. Isotonic solutions are those which have same osmotic pressure and hence same molar concentration.

We know that osmotic pressure, = i CRT where i = total no. of ions in electrolyte solution.

(i) BaCl2 Ba2+ + 2Cl– i = 1 + 2 = 3 1 ion 2 ion

= 3 × 0.2 RT = 0.6 RT

Urea (0.2 M) does not ionise. So, = CRT = 0.2 RT Since 0.6 RT and 0.2 RT differ, these are not isotonic solutions.

(ii) 0.1 M Urea. Urea does not ionise. So, = CRT = 0.1 RT 0.1 M NaCl. NaCl Na+ + Cl– i = 1 + 1 = 2 1 ion 1 ion

= i CRT = 2 × 0.1 RT = 0.2 RT Since 0.1 RT and 0.2 RT differ, these are not isotonic solutions.

(iii) 0.1 M NaCl. = 0.2 RT [see (ii) above]

0.1 M K2SO4. K2SO4 2K+ + SO 24 i = 2 + 1 = 3

2 ion 1 ion

= i CRT = 3 × 0.1 RT = 0.3 RT Since 0. RT and 0.3 RT differ, These are not isotonic solutions

(iv) 0.1 M Ba(NO3)2. Ba(NO3)2 Ba2+ + 2NO 3 i = 1 + 2 = 3

1 ion 2 ions

= i CRT = 3 × 0.1RT = 0.3 RT

0.1 M Na2SO4. Na2SO 2Na+ + SO 24 i = 2 + 1 = 3

2 ion 1 ions

= i CRT = 3 × 0.1 RT = 0.3 30. CNS– ions give red colour with Fe3+ ions in aqueous solution as:

)aq(3

)aq( CNS3Fe

red

)aq(3)CNS(Fe

If 0.1 M KCNS solution is separated from 0.1 M FeCl3 solution by means of semipermeable membrane, red colour will appear on:

a. FeCl3 solution side b. KCN’s solution side c. both sides d. neither side D Sol. Only solvent molecules and not the solute molecules or ions can pass through the semipermeable

membrane. Hence, Fe3+ and CNS– ions do not come in contact on either side

31. If is the degree of dissociation of Na2SO4, the Van’t Hoff’s factor (i) used for calculating the molecular mass is:

a. 1 + b. 1 - c. 1 + 2 d. 1 – 2 C

Sol. Na2SO4 2Na+ + SO 24

Initial mol 1 0 0

Mol after dissociation 1 - 2

Van’t Hoff’s factor, i = ondissociati before mol of no. Total

ondissociati after mol of no. Total



= 1

21 = 1 + 2 .

32. The Van’t Hoff factor for BaCl2 at 0.01 M concentration is 1.98. The percentage of dissociation of BaCl2 at this concentration is:

a. 49 b. 69 c. 89 d. 98 A

Sol. BaCl2 Ba2+ + 2Cl– Initial conc. 1 0 0

Conc. after dissociation 1- 2

Total no. of mol = 1 - + + 2 = 1 + 2

Hence Van’t Hoff factor i = 1 + 2 or = 2

1i

= 2

98.0

2

198.1

= 0.49

%age dissociation at 0.01 M concentration = 01.0

49.0= 49

33. Number of millitre of 0.25 M H2SO4 required to dissolve 2.1 g magnesium carbonate (at. wt., Mg = 24,

C = 12, O = 16) is: [MgCO3 + H2SO4 MgSO4 + CO2 + H2O]

a. 200 mL b. 100 mL c. 50 mL d. 20.0 mL B

Sol. 1000 mL of 1 M H2SO4 react with MgCO3 = g. mol. wt. of MgCO3 = 84 g

x mL of 0.25 M H2SO4 react with Mg CO3 = 3MgCO g 25.0x1000

84

So : 1000

25.0x84 = 2.1 g MgCO3 (given). Or x = 100 mL

34. H2S a toxic gas with rotten egg like smell is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195 m the Henry’s law constant will be

a. 282 bar b. 350 bar d. 145 bar d. 270 bar A Sol. Solubility of H2S = 0.195 m = 0.l195 mole in 1 kg of solvent

1kg of solvent (water) = 1000g = 18

1000 = 55.55 mole

Mole fraction of H2S gas in the solution = 55.55195.0

195.0

= 0.0035

Pressure at STP = 0.987 bar

According to Henry’s Law; S2HHS2H xKP

bar0035.0

987.0

x

PK

S2H

S2HH = 282 bar

35. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 200C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be:

a. 0.200 b. 0.549 c. 0.786 d. 0.478 D

Sol. Ratio of pentane, C5H12 and hexane, C6H14 i.e., .4

1

HC

HC

146n

125n



Hence, mole fraction ;5

1

41

1x

125HC

mole fraction ; mm 440P ;

5

4

41

4x 0

HCHC 125146

mm. 120P0HC 146

According to Raoult’s law:

146146125125 HC

0HCHC

0HC

xPxPPT

mm. 18496885

4120

5

1440

478.0184

88V etanpen

36. Some of the following gases are soluble in water due to formation of their ions: I: CO2 II: NH3 III: HCl IV: CH4 V: H2 Water-insoluble gases can be: a. I, IV, V b. I, V c. I, II, III d. IV, V D

37. Two liquids X and Y form an ideal solution. The mole fraction of X in vapour phase is 0.2 while in liquid phase is 0.8. If the total pressure is 800 torr then what are the original vapour pressure of X and Y?

a. 300PX 200PY b. 200PX 3200PY

c. 400PX 640PY d. 640PX 400PY

B Sol. Let mole fraction be x

Mole fraction of X in vapour phase = Total

X0X

P

xP

0.2 = 800

8.0P0X

0X

P = 200

Mole fraction of Y in vapour phase = 0.8 Mole fraction of Y in liquid phase = 0.2

Mole fraction of Y in vapour phase= 800

xP Y0Y

0.8 = 800

2.0PY

Hence XP = 200 and

YP = 3200

SECTION – B (Assertion-Reason Type) Negative Marking [-1]

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 × 4 = 20 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) Negative Marking [-1]

This Section contains 5 Questions. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 5 × 4 = 20 Marks

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

TA

XA = 1 X4 = .4

XB = .6

XB = 1

B A

C

TB



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. 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

Sol. Glycol have more number of moles due to less M.W 4. 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 5. 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 (More than One Answer) No Negative Marking

This Section contains 7 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. 7 × 5 = 35 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. 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 3. 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

4. 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

5. 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.

6. 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.

0.1M Na2 SO4

0.2M PbCl2

Semipermeable membrane

Vessel - 1 Vessel - 2 Vessel - 3

2C6H5COOH Benzene

C6H5 C

O H O

O H O

C – C6H5

– +

+ – Dimer



7. 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

SECTION – E (Matrix Type) No Negative Marking

This Section contains 3 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. 8 × 3 = 24 Marks

1. Match Column – I with Column – II (One or more than One Match)

Column – I Column – II

A. 1 : 0.1 M glucose; 2 : 0.1 M urea p. 1 and 2 are isotonic.

B. 1 : 0.1 M NaCl; 2 : 0.1 M Na2SO4 q. No net migration of solvent across the membrane

C. 1 : 0.1 M NaCl; 2 : 0.1 M KCl r. 1 is hypertonic to 2

D. 1: 0.1 M CuSO4; 2 : 0.1 M sucrose s. 1 is hypotonic to 2

Sol. A p, q; B s; C p, q; D r 2. Match Column – I with Column – II (More than one Match)

Column – I (Solutions) Column - II

(A) Benzene + Ethanol (p) High boiling azeotropes

(B) Benzene + Toluene (q) Vapour pressure is more than expected from Raoult’s law

(C) Acetone + Aniline (r) Can be separated by fractional distillation.

(D) HNO3 + H2O (s) Non-ideal solution

Sol. A q, s; B r; C p,s; D p,s 3. Match the entries of column I with appropriate entries of column II. (More than one match)

Column – I Column - II

(A) Al2(Cr2O7)3 Cr+3 (p) eq. Wt. = 18

Wt.M

(B) H2O2 O2 (q) Reduction Reaction

(C) Ba(MnO4)2 BaMnO4 (r) eq. Mass = 2

Wt.M

(D) As2O5 As2O3 (s) Oxidation reaction

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

SECTION – F (Integer Type) No Negative Marking

This Section contains 6 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. 6 × 4 = 24 Marks

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 6. Calculate the parts per million of SO2 gas in 250 mL water (density 1 g cm–3) containing 5 × 10–4 g of

SO2 gas. Sol. (2) Mass of SO2 gas = 5 × 10–4 g; Mass of H2O = Volume × Density = 250 cm3 × 1 g cm–3 = 250 g

Parts per million of SO2 gas = g 250

g 105 4× 106 = 2 Ans.

1 atm

(1) (2)

2 atm

2 atm

(3)

2 atm

(5)

2 atm

(4)

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