Chapter 10

Solutions
Solutions

SECTION - A
Objective Type Questions (One option is correct)

[Concentration Term and Raoult’s Law]

1. What is the molarity of 4.9% H3PO4 solution by mass (density of H3PO4 = 1.22 g/ml)?
(1) 0.61 M (2) 4.9 M (3) 1.22 M (4) 1 M
Sol. Answer (1)
Given:  of H3PO4 solution = 1.22 g/ml
Let the volume of solution = 1000 ml
m
  mSolution = 1.22 × 1000 = 1220 g
v
4.9
WH3PO4 =  1220 g
100
W 4.9  1220 4.9  1220
n 
M 100  980 1000  98
W 4.9 1220
 n   0.61
M 100 98
 0.61 moles are present in 1L
 Molarity is 0.61 M

2. The molality of 1 M NaNO3 solution is (d = 1.25 g/ml)

(1) 0.8 m (2) 0.858 m (3) 1.6 m (4) 1 m
Sol. Answer (2)
For NaNO3 solution, C = 1 M
m
=  mSolution =  × V
V
 mSolution = 1.25 × 1000 =1250 g

WNaNO3 = 1 × (23 + 14 + 48) = 85

WSOLVENT = (1250 – 85) = 1165 g

1 103
 Molality = = 0.858 m
1165

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Solutions of Assignment Solutions 35
3. The solubility of a gas in a liquid generally increases with
(1) Increase in temperature (2) Amount of liquid taken
(3) Decrease in temperature (4) Reduction of gas pressure
Sol. Answer (3)
Dissolution process is exothermic

4. When a saturated solution of sodium chloride is heated it

(1) Remains saturated (2) Becomes unsaturated
(3) Becomes supersaturated (4) Achieves equilibrium state
Sol. Answer (2)
NaCl gets precipitated out.

5. The vapour pressure of pure liquid A is 70 torr at 27°C. It forms an ideal solution with another liquid B. The
mole fraction of B is 0.2 and total vapour pressure of the solution is 84 torr at 27°C. The vapour pressure of
pure liquid B at 27°C is
(1) 140 torr (2) 50 torr (3) 14 torr (4) 70 torr
Sol. Answer (1)
Given: PoA = 70 torr

XB = 0.2; PT = 84 torr

PoB = ?; XA = 1 – XB = 1 – 0.2 = 0.8

We know that

PT = PoA XA + PoB XB

 (84) = (70) (0.8) + PoB (0.2)

28
or = PoB = 140 mm Hg
0 .2

6. If two liquids A and B form an ideal solution, then

(1) Enthalpy of mixing is zero (2) Entropy of mixing is zero
(3) Free energy of mixing is zero (4) Both free energy and entropy of mixing is zero
Sol. Answer (1)
Hmix = 0 but Smix and Gmix is not zero.

7. Raoult’s law is not valid in

(1) Non-ideal solution showing positive deviation (2) Non-ideal solution showing negative deviation
(3) Mixture of two immiscible liquid (4) All of these
Sol. Answer (4)
Non-ideal solution do not obey Raoult’s law, both negative and positive but it is valid for both negative and
positive non-ideal solution.
Non-ideal solutions are the part of miscible liquid.
Raoult’s law is valid for miscible liquid only it is not for immiscible liquid.

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36 Solutions Solutions of Assignment

[Colligative Properties (Non-electrolytic solute)]

8. 20 g of non-electrolyte, non-volatile solute (CxH2xOx), when dissolved in 100 gm water at 100°C, lowers the
1
vapour pressure of solution by th of the vapour pressure of pure water at this temperature. What is formula
100
of the compound?
(1) C6H12O6 (2) C12H24O12 (3) C44H88O44 (4) C3H6O3
Sol. Answer (2)
Wsolute = 20 g ; Wsolvent = 100 g
For pure water, P° is the vapour pressure.

P 99P
PS = P° – =
100 100
We know that

20
P 1 M
 
100 P 20 100

M 18

 M = 360
 Formula of non-electrolyte is C12H24O12

9. The vapour pressure of a solvent decreases by 5.4 torr when a non-volatile solute is added. In this solution,
mole fraction of solute is 0.2. What would be mole fraction of the solvent if decrease in vapour pressure is
16.2 torr?
(1) 0.6 (2) 0.4 (3) 0.2 (4) 0.8
Sol. Answer (2)
(P)1 = P° – PS = 5.4 torr
Xsolute = 0.2
(P2) = 16.2 torr;
We know that

(P  PS )1
P  Xsolute 1 0.2
 
(P  PS )2  Xsolute  x
2
P

5.4 0.2 0.2  16.2

  x =  Xsolute = 0.6  Xsolvent = 1 – 0.6 = 0.4
16.2 x 5.4

10. When NaCl is added to aqueous solution of glucose

(1) Freezing point is raised (2) Freezing point is lowered
(3) Boiling point is lowered (4) No change in freezing point or boiling point
Sol. Answer (2)
NaCl is a solute which will cause the depression in the freezing point as it is one of the colligative property.

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Solutions of Assignment Solutions 37
11. 2.56 g of sulphur in 100 g of CS2 has depression in freezing point of 0.010°C Kf = 0.1°C (molal)–1. Hence
atomicity of sulphur in the solution is
(1) 2 (2) 4 (3) 6 (4) 8
Sol. Answer (4)
WSOLUTE = 2.56 g
WSOLVENT (CS2) = 100 g
Tf = 0.01° C ; Kf = 0.1
We can apply the equation,
Tf = Kf xm
(0.1)  2.56  1000
(0.01) =
M  100
2.56  1000  0.1
M= = 256 g
100  0.01
 No. of atoms of Sulphur
256
 =8
32
 Atomicity = 8

12. ‘A’ gram of non-volatile, non-electrolyte (molar mass M) is dissolved in 200 ml of unknown solvent (density = 1.25
gm/ml molal elevation constant is Kb). Elevation in boiling point of this solution can be given by
M 4 KbA KbA K bM
(1) K (2) (3) (4)
b M 4M 4A
Sol. Answer (2)
Wsolute = A ;
Vsolvent = 200 ml ;  = 1.25 g/ml
Tb = Kb × molality
A 1000 4A
Molality = M  200  1.25 =
M
4K b A
So, Tb =
M

13. 75 g ethylene glycol is dissolved in 500 gram water. The solution is placed in a refrigerator maintained at a
temperature of 263.7 K. What amount of ice will separate out at this temperature?
(Kf water = 1.86 K molality–1)
(1) 300 g (2) 200 g (3) 178 g (4) 258 g
Sol. Answer (4)
W ( CH2OH.CH2OH) = 75 g
W (H2O) = 500 g and Tf = 9.3
75  1000  1.86
 (9.3) = 62  Wsolvent

75  1000  1.86
 W solvent = = 242 g
62  9.3
Ice separated = 500 – 242 = 258 g
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38 Solutions Solutions of Assignment

[Abnormal Colligative Properties]

14. If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol of H2O, then %
ionization of NaCl is
(1) 60% (2) 80% (3) 40% (4) 100%
Sol. Answer (4)
P  Ps in

P in  N
i1
0.4 =
i 1 3
i=2
For NaCl, i = 1 + 
or 2 = 1 + 
or  = 1
 NaCl is 100% ionized.

15. Amongst the following, the solution which shows highest osmotic pressure is
(1) 0.05 M NaCl (2) 0.10 M BaCl2
(3) 0.05 M FeCl3 (4) 0.05 M Na2SO4
Sol. Answer (2)
Osmotic Pressure  = i×C×S×T
NaCl (i = 2) C1 = 0.05
BaCl2 (i = 3) C2 = 0.10
FeCl3 (i = 4) C3 = 0.05
Na2SO4 (i = 3) C4 = 0.05
 Product of C and i is maximum for BaCl2.
 BaCl2 solution has maximum Osmotic pressure.

16. A 0.2 molal aqueous solution of weak acid HX is 20% ionized. The freezing point of solution is
(Kf = 1.86)
(1) – 0.45°C (2) – 0.9°C (3) – 0.31°C (4) – 0.53°C
Sol. Answer (1)
Weak acid HX is 20% ionized

  –
 H X
HX 
1–   

i = 1 –  +  +  =1 +  = 1 + 0.2 = 1.2
Tf = (0.2) × (1.86) × (1.2) = 0.45°C
 Tf = 0 – 0.45°C = – 0.45°C

17. When 20 g of napthanoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg/mol) a freezing point
depression of 2 K is observed. The van’t Hoff factor (i) is
(1) 0.5 (2) 2.0 (3) 1.0 (4) 3.0

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Solutions of Assignment Solutions 39
Sol. Answer (1)
W = 20 g ; Wsolvent =50 g
Kf = 1.72 ; Tf = 2K
We can use the equation
Tf = i × Kf   m
ΔTf 2  172  50
 i = K  m  1.72  20  1000
f

 i = 0.5

18. van’t Hoff factor of aqueous acetic acid solution is

(1) < 1 (2) > 1 (3) = 1 (4) 0
Sol. Answer (2)
CH3COOH undergoes ionization in aqueous solution therefore i > 1.


 CH3COO– + H+
CH3COOH 

19. Acetic acid dimerises in benzene solution. The van’t Hoff factor for the dimerisation of acetic acid is 0.8. The
% of dimerisation of acetic acid is
(1) 20% (2) 40% (3) 60% (4) 80%
Sol. Answer (2)


 (CH3COOH)2
For, 2 CH3COOH 
i = 1 –  + 0.5 
0.8 = 1 – 0.5 
0.5  = 0.2
 = 0.4
Hence % dimerisation is 40.

20. Which has lowest osmotic pressure at same temperature?

(1) 100 ml of 1 M urea (2) 400 ml of 0.5 M KCl
(3) 200 ml of 0.5 M glucose (4) 300 ml of 0.25 M K3PO4
Sol. Answer (3)
For 0.5 M C6H12O6 the osmotic pressure will be least because i = 1
For urea,  = 1 × 1 × ST
For KCl,  = 2 × 0.5 × (S)(T) = ST
For K3PO4,  = 4 × 0.25 × ST = ST
Hence, osmotic pressure will be least for 0.5 M C6H12O6.

21. The value of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The
degree of dissociation of silver nitrate is
(1) 60% (2) 83.5% (3) 46.7% (4) 60.23%
Sol. Answer (2)
Colligative properties are inversely related to molecular mass of the solute. Because of which
MCAL 170
i=  = 1.83
MOBS 92.64

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40 Solutions Solutions of Assignment

Since
i = 1 + (n – 1) 
For AgNO3, i=1+
  = (1.83 – 1) = 0.835
 83.5% ionization takes place

22. If a solute undergoes dimerisation and trimerisation, the minimum values of the van’t Hoff factors are
(1) 0.5 and 1.50 (2) 1.5 and 1.33 (3) 0.5 and 0.33 (4) 0.25 and 0.67
Sol. Answer (3)


2A 
 A2 (Dimerization)
 
i=1–+ = 1
2 2
1 1
For minimum ,  = 1  i = 1 
2 2
or i = 0.5

For trimerization 


 A3
3A 
 ⎛1 ⎞ ⎛ 2 ⎞
i=1–+ = 1+  ⎜ 3  1⎟ = 1+  ⎜ 3 ⎟
3 ⎝ ⎠ ⎝ ⎠
2 2 1
 i=1– = 1 = = 0.33
3 3 3

23. 2 millimolar solution of sodium ferrocyanide is 60% dissociated at 27°C. Osmotic pressure of the solution is
(1) 2.14 atm (2) 1.02 atm (3) 0.167 atm (4) 0.0234 atm
Sol. Answer (3)
Na4 [Fe(CN)6] undergoes 60% ionization
i = 1 + (n – 1)  = 1 + (5 – 1)  = 1 + 4  = 1 + 4(0.6) = 3.4
C = 2 × 10–3M
 =i×C×S×T
  = (3.4) (2 × 10–3) (0.0821) (300) = 0.167 atm.

24. A water sample contains 9.5% MgCl2 and 11.7% NaCl (by weight). Assuming 80% ionisation of each salt.
Boiling point of water will be approximately (Kb = 0.52)
(1) 110.01°C (2) 377 K (3) 277.25 K (4) 102.5°C
Sol. Answer (2)

25. A complex is written as M(en)y.xBr. Its 0.05 molar solution shows 2.46 atm osmotic pressure at 27°C.
Assuming 100% ionisation and coordination number of metal (III) is six, complex may be
(1) [M(en)2Br2]Br (2) [M(en)3]Br3 (3) [M(en)2Br]Br2 (4) [M(en)]Br3
Sol. Answer (1)
The complex is M (en)Y . x Br
C = 0.05M ;  = 2.46 atm ; T = 300 K
Osmotic pressure of the solution is given by
=i×C×S×T

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Solutions of Assignment Solutions 41

2.46
 i = =2
(0.05)(0.0821)(300)
The vant-Hoff factor is 2 and since the coordination no. is 6 and n = 2 ; Formula is [M(en)2 Br2] Br
On ionization it forms two ions

[M(en)2 Br2] Br   [M(en)2 Br2]+ + Br–

26. Consider three solutions of 3 strong electrolytes. AB, CD2 and EF3

2M 3M 4M

AB solution CD2 solution EF3 solution

at 127°C at 438.11°C at 27°C
(I) (II) (III)
The osmotic pressure ratio of I, II, and III is
(1) 1 : 4 : 3 (2) 1 : 2 : 3 (3) 2 : 3 : 4 (4) 2 : 4 : 3
Sol. Answer (1)
For (I)
AB  A+ + B–
1 = i × C1 × S × T1
= 2 × (2) × 400 × S = 1600 S
For (II)
CD2  2D– + C2+
2 = 3 × 3 × 711.11 × S = 6400 S
For (III)
EF3  E3+ + 3F–
3 = 4 × 4 × 300 × S = 4800 S
 Ratio of Osmotic pressures is
1600 S : 6400 S : 4800 S / i.e., 1 : 4 : 3

27. Degree of dissociation of three binary electrolytes AB, CD and EF are 60%, 20% and 100% in the solution
having same mole fraction of water. Ratio of lowering in vapour pressure of their solution is
(1) 0.8 : 0.6 : 1 (2) 0.2 : 0.4 : 0.1 (3) 0.3 : 0.5 : 0.2 (4) 1 : 2 : 0.5
Sol. Answer (1)
i for AB = 1 + 0.6 = 1.6
for CD ; i = 1.2
and for EF ; i = 1 + 1 =2
(P)1 = i P° . (Xsolute)1
(P)2 = i P° . (Xsolute)2
(P)3 = i P° . (Xsolute)3
(1.6)P° : 1.2P° : 2P°
1.6 : 1.2 : 2
0.8 : 0.6 : 1
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42 Solutions Solutions of Assignment

28. Osmotic pressure of a blood sample is 4.92 atm at 27°C. Which of the following is not isotonic with blood
sample?
(1) 3.6% (m/m%) glucose aqueous solution (2) 1.2% (m/m%) urea aqueous solution
(3) 0.585% (m/m%) NaCl aqueous solution (4) 1.7% (m/m%) NaNO3 aqueous solution
Sol. Answer (4)
1.7% NaNO3 solution has the concentration as

⎛ 1.7 ⎞
2⎜ ⎟  10  0.0821 300  9.852 atm
⎝ 85 ⎠ (S) (T)

which is not isotonic with blood sample.

29. 0.067 molar aqueous solution of a binary electrolyte A+B– shows 2.46 atm osmotic pressure at 27°C. What
fraction of A+B– remains unionised?
(1) 10% (2) 15% (3) 50% (4) Zero
Sol. Answer (3)
For AB ; C = 0.067 M
 = 2.46 atm ; T = 300 K


i=
CST

2.46
i = 1.5
0.067  0.0821 300

For electrolyte AB
i=1+   = 0.5 or 50% ionized

30. The value of observed molecular weight of silver nitrate is 132.5 gram/mole, in an aqueous solution. The degree
of dissociation of silver nitrate in this solution may be
(1) 79% (2) 32% (3) 28% (4) 44%
Sol. Answer (3)

MCAL
i = M
OBS

MAgNO3 170
i=  = 1.28
132.5 132.5
For AgNO3
i = 1 + (n – 1)  = 1 +  = 1.28
 = 0.28
Hence, 28% ionization takes place.

31. When mercuric iodide is added to an aqueous solution of KI the

(1) Boiling point increases (2) Boiling point decreases
(3) Freezing point decreases (4) Osmotic pressure increases

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Solutions of Assignment Solutions 43
Sol. Answer (2)
When HgI2 is added to KI it forms the complex which indicates that no. of solute particles decreases.



 K2[HgI4]
HgI2 + KI 
 Boiling point decreases.

32. Which of the following aqueous solutions has the highest freezing point (assuming m = M)?
(1) 0.1 M KNO3 (2) 0.2 M Na3PO4 (3) 0.25 M FeCl3 (4) 0.01 M Na2SO4
Sol. Answer (4)
For KNO3
(Tf)1 = 2 × 0.1 × m = 0.2 m
For Na3PO4
(Tf)2 = (0.2) × (4) × m = 0.8 m
For FeCl3
(Tf)3 = (0.25) × 4 × m = m
For Na2SO4
(Tf)4 = (0.01) × 3 m = 0.03 m
Hence, maximum depression in freezing point is for Na3PO4, and least for Na2SO4
Hence, F.P. is highest for Na2SO4 solution.

33. Which of the following pair of aqueous solutions can be expected to be isotonic at the same temperature?
(Consider each electrolyte according to its 100% dissociation)
(1) 0.1 M urea and 0.1 M NaCl (2) 0.1 M NaCl and 0.1 M Na2SO4
(3) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4 (4) 0.1 M glucose and 0.2 M MgCl2
Sol. Answer (3)
0.1 M Ca(NO3)2 solution has
1 = (0.1) (3) × ST = (0.3) ST …….. (i)
and Na2SO4 solution
2 = (0.1) (3) × ST = (0.3) ST …….. (ii)
Hence, 1 = 2 which means that both the solutions will be isotonic which means both have same osmotic
pressure.

34. Among the following solutions

a. 0.01 M NaCl b. 0.05 M glucose
c. 0.01 M CaCl2 d. 0.02 M KCl
The correct order of decreasing boiling point can be given as (assuming same dissociation)
(1) a > b > c > d (2) b > d > c > a (3) a > c > d > b (4) d > c > b > a
Sol. Answer (2)
Tb = i × kb × m
For glucose, i = 1 as  = 0

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44 Solutions Solutions of Assignment

35. When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a freezing
point depression of 2 K is observed. The van’t Hoff factor (i) is [IIT-JEE 2007]
(1) 0.5 (2) 1 (3) 2 (4) 3
Sol. Answer (1)
Tf = i × kf × m
20 1000
2 = i × 1.72 × 
172 50
i = 0.5

36. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of
N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure
is [IIT-JEE 2009]
(1) 4.0 × 10–4 (2) 4.0 × 10–5 (3) 5.0 × 10–4 (4) 4.0 × 10–6
Sol. Answer (1)
As the mole fraction of N2 gas is 0.8 and total pressure is 5 atm. Therefore the partial pressure of N2 will
be 4 atm.
Now Pa = KH Ha

n
4 = 1 × 105  (∵ n is much less than 10)
n  10

n = 4 × 10–4 moles

37. The freezing point (in °C) of a solution containing 0.1 g of K3[Fe(CN)6] (Mol. wt. 329) in 100 g water
(Kf = 1.86 K kg mol–1) is [IIT-JEE 2011]
(1) –2.3 × 10–2 (2) –5.7 × 10–2 (3) –5.7 × 10–3 (4) –1.2 × 10–2
Sol. Answer (1)

0.1
329
Tf – Tf = 0 – Tf = 4 × 1.86 × 100
1000
0.1 1000
Tf   4  1.86    0.023
329 100

38. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in
boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration
of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol–1) [IIT-JEE 2012]
(1) 724 (2) 740 (3) 736 (4) 718
Sol. Answer (1)
Tb = 1 × Kb × m
 2 = 1 × 0.76 × m

2
 m=  2.63
0.76

760  PS 18
  2.63 
760 1000
 PS = 724 mmHg

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Solutions of Assignment Solutions 45
39. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing
point of the solution. Use the freezing point depression constant of water as 2 K kg mol–1. The figures shown
below represent plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is
46 g mol–1]
Among the following, the option representing change in the freezing point is [JEE(Advanced)-2017]

1
Wate ter
r Wa
Ice 1 nol
V.P./bar

V.P./bar
tha
(1) (3) Ice +E
ter
Wa
Water + Et
h anol
270 273 271 273
T/K T/K

1
ter Wate
Wa r
1 nol Ice
V.P./bar

V.P./bar
tha
(3) Ice +E (4)
ter
Wa
Water + Et
h anol
270 273 271 273
T/K T/K
Sol. Answer (3)

⎡ W  1000 ⎤
Tf = iK f ⎢ 2 ⎥
⎣ M2  W1 ⎦

⎡ 34.5  1000 ⎤
= 1 2 ⎢
⎣ 46  500 ⎥⎦
=3K
273 (K) – Tf = 3 (K)
 Tf = 270 K
Also, with decrease in temperature, V.P. decreases.
 Graph (C) is correct.

SECTION - B

Objective Type Questions (More than one options are correct)

1. 100 ml 20%(by mass) H 2SO 4 (density = 1.2 gm/ml) and 100 ml 40% (by mass) H 2SO 4 (density =
1.4 gm/ml) are mixed together. Which are the correct concentration terms for this mixture?
(1) Molality = 2.54 (2) Molarity = 2.04
(3) Molality = 4.54 (4) Molarity = 4.08
Sol. Answer (3, 4)
 = 1.2 g/ml ; V = 100 ml and 20% by mass is H2SO4

m 1.2
=  m =  × V = × 100 ml = 120 g
V ml
20
 W H2SO4 = × 120 = 24 g
100

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46 Solutions Solutions of Assignment

Secondly;  = 1.4 g/ml; V = 100 ml = 40% by mass

m
 =  m = 1.4 × 100 = 140 g
V

40
 WH2SO4 = × 140 = 56 g
100

 On mixing both solutions

Final Molarity is

24 64
 moles in 200 ml
98 98

24  56
 In 1000 ml  ×5
98

80
 × 5 = 4.08
98

 Final molarity is 4.08 M

24  56
Moles of H2SO4 is
98

WSOLVENT = (120 – 24) + (140 – 56) = 180 g

80
 In 180 g SOLVENT  MOLES (H2SO4) =
98

For 1000 g SOLVENT

80
× 1000 = 4.54 m
98  180

2. Which are the correct statements?

(1) CHCl3 + CCl4 – Endothermic solution
(2) Acetic acid + Pyridine – Hot solution
(3) HNO3 + Water – Endothermic solution
(4) Water + HI – Minimum boiling point azeotrope
Sol. Answer (1, 2)
It is a known fact that CHCl3/CCl4 forms endothermic solution as H > 0.
CH3COOH and pyridine forms a hot solution due to negative deviation from Raoult’s Law.

3. V litre decinormal solution of NaCl is prepared. Half of the solution is converted into centinormal and added
to the left decinormal solution. Then
1
(1) Number of millimoles of NaCl are reduced by
5
(2) Number of milliequivalents of NaCl do not change
(3) Normality of the final solution becomes 0.01 N
(4) Molarity of the final solution becomes 0.018 M
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Solutions of Assignment Solutions 47
Sol. Answer (2, 4)
Given:

N
Initial volume = V ; Normality =
10

V 1 V
Equivalents = =
10 10

50% solution is taken

V 1 V  1
 =
 10 100

100  V
V  = 5V
20

Final volume = (5V)

Now there are two solutions

I solution (Left solution) II-Prepared solution

V N N
; 5V ;
2 10 100

 On mixing

N1V1  N2 V2 ⎛ V V ⎞ ⎛V ⎞
  5V ⎟
N = (V1  V2 ) = ⎜⎝ 20 20 ⎟⎠ ⎜⎝ 2 ⎠

V 2 2
 N =  =
10 11V 110
= 0.018 N  0.018 M
And no. of equivalent does not change on dilution.

4. At 300 K, the vapour pressure of an ideal solution containing 1 mole of A and 3 moles of B is
500 mm Hg. At the same temperature, 2 moles of B are added to this solution. The vapour pressure of solution
increases by 10% of the original vapour pressure. Correct statements about the vapour pressure are
(1) Vapour pressure of A in the pure state is 50 mm Hg
(2) Vapour pressure of B in the pure state is 650 mm Hg
(3) Ratio of final vapour pressure to the initial vapour pressure is 1 : 0.5
(4) Ratio of vapour pressure of pure B to the vapour pressure of pure A is 13 : 1
Sol. Answer (1, 2, 4)
nA = 1 ; nB = 3 ; PT = 500 mmHg
When ‘2’ moles of ‘B’ is added

nB = (3 + 2) = 5 ; nA = 1

500  10
PT = 500 + = 50 + 500 = 550
100

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48 Solutions Solutions of Assignment

By equation:

⎛ 1 ⎞ ⎛ 3 ⎞
500 = PA ⎜ ⎟  PB ⎜ 1  3 ⎟
⎝ 1  3 ⎠ ⎝ ⎠

(500 × 4) = PA  3PB … (i)

Secondly we can write

⎛ 1⎞ ⎛5⎞
550 = PA ⎜ ⎟  PB ⎜ ⎟
⎝6⎠ ⎝6⎠

 PA  5PB = 6 × 550 … (ii)

Subtracting (i) from (ii)

1300
2PB = 3300 – 2000 = 1300  PB = = 650 mmHg
2

and PA = 2000 – (3 × 650)  PA = 2000 – 1950 = 50 mm Hg

PB 650 13
and = =
PA 50 1

5. Consider the following graphs

V V
V
Temperature

Temperature

V L L
L L

A = 1 B = 1 A = 1 B = 1
B = 0 A = 0 B = 0 A = 0
Graph-I Graph-II

Choose the correct statements

(1) According to both graphs mole fraction of A > mole fraction of B in condensate
(2) Graph I belongs to minimum boiling azeotrope
(3) Graph II belongs to maximum boiling azeotrope
(4) Graph II belongs to minimum boiling azeotrope while graph I belongs to minimum boiling azeotrope
Sol. Answer (1, 2, 3)
Graph (I) and graph (II) are the graphs of azeotropes and is clear from the above vapour-liquid equilibrium
that
(1) is MINIMUM BOILING AZEOTROPES and
(2) is MAXIMUM BOILING AZEOTROPES
and it is also clear that XA > XB in the condensate formed from the graph.

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Solutions of Assignment Solutions 49
6. The following graph is plotted between the vapour pressures of two volatile liquids against their respective mole
fractions

A = 1 A = 0
B = 0 B = 1
X

Which of the following combinations are correct?

(1) When A = 1, P = PA° (2) When B = 1, P > PA°
(3) When A = 1, P < PA° (4) When B = 1, P = PB°
Sol. Answer (1, 2, 4)
Fact.

7. A binary liquid (AB) shows positive deviation from Raoult’s law when

(1) PA  PAo liquid

A and PB  PBo Bliquid (2) Intermolecular forces A – A, B – B > A – B

(3) Vmix > 0 (4) Hmix > 0

Sol. Answer (1, 2, 3, 4)
Fact.

8. In the depression of freezing point experiment, it is found that

(1) Vapour pressure of solution is less than that of pure solvent
(2) Vapour pressure of solution is more than that of pure solvent
(3) Only solute molecules solidify at the freezing point
(4) Only solvent molecules solidify at the freezing point
Sol. Answer (1, 4)
Addition of non-volatile solute lowers the vapour pressure of solvent.

9. At 27°C, the osmotic pressure of a non-volatile non-electrolyte solute is 47.7 mm Hg. The solution is diluted
and heated upto 127°C, when the osmotic pressure becomes 19 mm Hg. Then
(1) Molality of solution becomes half
(2) Number of milli moles of solute remains same
(3) Extent of dilution of solution is 3.4 times
(4) Relative lowering in vapour pressure reduces
Sol. Answer (2, 3, 4)
Given :  = 47.7 mmHg
Solution is diluted and is heated upto 127°C,  = 19 mmHg
Hence, on dilution, no. of millimoles remains same and relative lowering in vapour pressure decreases.

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50 Solutions Solutions of Assignment

10. Equimolal solutions of NaCl, BaCl2 and Na3PO4 are prepared in water, then correct statements are
(1) Freezing point of NaCl solution is –2°C if freezing point of BaCl2 solution is –3°C
(2) Freezing point of Na3PO4 is –4°C if freezing point of BaCl2 is –3°C
(3) Elevation in boiling point for BaCl2 solution is 1.5 times higher than NaCl solution
(4) Elevation in boiling point for NaCl is half of Na3PO4 solution
Sol. Answer (1, 2, 3, 4)
‘m’ for NaCl; BaCl2– and Na3PO4 is same
i (for NaCl) = 2
i (for BaCl2) = 3
(Tf)1 = 2 × Kf × m
(Tf)2= 3 × Kf × m
Hence (1) is correct answer
For Na3PO4 (i = 4) and for BaCl2 (i = 3)  freezing point is – 4°C and – 3°C
For BaCl2 (i = 3) and for NaCl (i = 2)
 Elevation in boiling point for BaCl2 is 1.5 times higher than NaCl solution
and for NaCl (i = 2)
Na3PO4 (i = 4)
 Elevation in BP for NaCl is half of Na3PO4 solution

11. Choose the pairs having identical value of van’t Hoff factor
(1) 0.05 M K4[Fe(CN)6] (50% degree of dissociation) and 0.05 M Mohr salt (80% degree of dissociation)
(2) 0.2 M NaCl (80% degree of dissociation) and 0.2 M BaCl2 (40% degree of dissociation)
(3) 0.05 M Na3PO4 (60% degree of dissociation) and 0.05 M K4[Fe(CN)6] (45% degree of dissociation)
(4) 0.01 M NaNO3 (90% degree of dissociation) and 0.01 M FeCl3 (30% degree of dissociation)
Sol. Answer (2, 3, 4)


 Na+ + Cl–
NaCl 

i = 1 + (2 – 1) (0.8) = 1.8



 Ba2+ + 2Cl–
BaCl2 

i = 1 + (3 – 1) (0.4) = 1.8
 (2) is the correct answer because both have identical values of i.



 3Na+ + PO43–
Na3PO4 

i = 1 + (3) (0.6) = 2.8



 4K+ + [Fe(CN)6]4–
K4 [Fe(CN)6] 

i = 1 + (4) (0.45)
i = 1 + 1.8  2.8
(3) is correct answer

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Solutions of Assignment Solutions 51


and NaNO3  +
 Na + NO3
–

1 + (2 – 1)  = 1 +  = 1 + 0.9 = 1.9
and for


 Fe3+ + 3Cl–
FeCl3 
1 + (4 – 1)  = 1 + 3
= 1 + 3 (0.3) = 1.9
So, (4) is the correct answer.

12. Consider the following arrangement and choose the correct options

0.2 M 0.15 M
Na2SO4 KCl

Semipermeable membrane

(1) O.P. of Na2SO4 solution is lesser than the O.P. of KCl solution
(2) Water will flow from KCl solution to Na2SO4 solution
(3) Water will flow from Na2SO4 solution to KCl solution
(4) O.P. of Na2SO4 solution is higher than the O.P. of KCl solution
Sol. Answer (2, 4)
For left hand compartment
0.2M Na2SO4; 1 = (0.2) × (S) × (T) × 3
1 = (0.6) (S) (T)
and for 0.15 M KCl
2 = (2) × (S) × (T) × (0.15)
= (0.3) (ST)
 1 > 2  Na2SO4 solution have higher osmotic pressure and water will flow from lower concentration
to higher concentration.
 Water will flow from RHC to LHC.
13. Which of the following pair of solutions can be expected to be isotonic at the same temperature?
(1) 0.1 M urea & 0.1 M NaCl
(2) 0.1 M urea & 0.1 M glucose
(3) 500 ml 0.2 M NaCl & 200 ml 0.2 N KCl
(4) 100 ml 0.05 N Ca(NO3)2 & 100 ml 0.15 M Na2SO4
Sol. Answer (2, 3)
Isotonic solutions have same osmotic pressures.
0.1M UREA ; 0.1M C6H12O6
Same concentration and i = 1 for both
For NaCl and KCl n-factor is 1
 0.2 M NaCl and 0.2 N KCl have same osmotic pressure.

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52 Solutions Solutions of Assignment

14. A compound X undergoes 100% pentamerisation in a given solvent Y. Correct statements are
(1) van’t Hoff factor of compound is 0.20
(2) Experimental elevation in boiling point
Calculated elevation in boiling point
=
5
(3) Observed molar mass of solute
Normal molar mass
=
5
(4) Observed freezing point × 5 = Normal freezing point
Sol. Answer (1, 2)
For pentamerization


5A 
 A5

 ⎛1 ⎞ 5  4
i=1–+ = 1 +  ⎜  1⎟ =
5 ⎝5 ⎠ 5

1
0.2 = i = ; ( = 1)
5
 van’t-Hoff factor is equal to 0.2 and 5 × (Experimental elevation in Boiling point) = Calculated elevation
in B.P.

15. 100 ml of 3% (weight/volume) urea solution and 100 ml of 6.84% (weight/volume) of cane sugar are mixed at
20°C. The solution is heated upto 27°C after mixing 100 ml water into it. Correct statements for final solution
are
(1) O.P. of solution is 4 atm
(2) O.P. of urea in the solution is 12.13 atm
(3) O.P. of cane sugar in solution is 1.64 atm
(4) O.P. of urea in the solution is 4 atm
Sol. Answer (3, 4)
UREA solution is 3% (w/v)

3 1
 nUREA = = moles = 0.05 mole in100 ml
60 20

6.84
and nC12H22O11 = = 0.02 mole in 100 ml
342

VT = 300 ml

(0.05  10) ST 0.05  10  0.0821  300

 (UREA) = = = 4 atm
3 3

0.02  10  0.0821  300

p (C12H22O11) = = 1.64 atm
3

So, Total O.P. = 4 + 1.64 = 5.64 atm

Hence, (3) & (4) options are correct.

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Solutions of Assignment Solutions 53
16. Correct statements about the following solutions (Assuming equal amount of each solute is dissolved in 1 L
solution)
1 L NaCl, 1 L (NH2)2CO, 1 L BaCl2, 1 L Ca3(PO4)2
(1) Osmotic pressure :
NaCl > (NH2)2CO > Ca3(PO4)2 > BaCl2
(2) Boiling point :
NaCl > (NH2)2CO > Ca3(PO4)2 > BaCl2
(3) Freezing point :
NaCl > (NH2)2CO > Ca3(PO4)2 > BaCl2
(4) Freezing point :
(NH2)2CO < NaCl < Ca3(PO4)2 < BaCl2
Sol. Answer (1, 2)
For (1)
NaCl (i = 2)
Urea (i = 1)
Ca3(PO4)2 (i = 5)
BaCl2(i = 3)
 Osmotic pressure and elevation in boiling point, both are colligative properties and order is same for both
the cases.

17. Choose the correct statements

(1) 1 M H2SO4 (d = 1 gm/ml) is more concentrated than 1 m H2SO4 (d = 1 gm/ml)
(2) Molality of solution is 1.136 if 2 gram-equivalents of H2SO4 is dissolved into 90.2 gm water
(3) Vapour pressure of solution becomes higher than ideal solution if there is a positive deviation (according
to Raoult’s law)
(4) When 0.1 M K4[Fe(CN)6] solution and 0.1 M FeCl3 solution is separated by a semi permeable membrane,
water flows from K4[Fe(CN)6] solution to FeCl3 solution
Sol. Answer (1, 3)
Vapour pressure of the solution becomes higher than ideal solution when there is positive deviation from
Raoult’s law.
And 1 M solution is always more concentrated then 1 molal solution because molarity is w.r.t. volume of
solution, but molality is defined w.r.t. weight of solvent.

18. Dimer of acetic acid in benzene is in equilibrium with acetic acid at a particular condition of temperature and
pressure. If half of the dimer molecules are hypothetically separated out then
(1) Osmotic pressure of the solution reduces (2) Freezing point of the solution reduces
(3) Boiling point of the solution reduces (4) Vapour pressure of the solution reduces
Sol. Answer (1, 3)
Acetic acid undergoes dimerization as


2CH3COOH 
 (CH3COOH)2
Since i < 1,  Osmotic pressure decreases and boiling point also decreases as  and (  Tb) both are
colligative properties.

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54 Solutions Solutions of Assignment

19. In which of the following pair of solutions will the values of van’t Hoff factor be same?
(assuming same dissociation)
(1) 0.05 M K4[Fe(CN)6] and 0.10 M FeSO4
(2) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4(NH4)2 SO4.6H2O
(3) 0.20 M NaCl and 0.10 M BaCl2
(4) 0.05 M FeSO4.(NH4)2SO4.6H2O and 0.02 M KCl. MgCl2.6H2O
Sol. Answer (2, 4)
Number of particles dissociated are same

20. Choose correct option about given figure

SPM
0.1 M 0.05 M
NaCl BaCl2
Solution Solution

(1) No movement of liquid from any solution across SPM

(2) BaCl2 will flow towards NaCl solution
(3) NaCl will flow towards BaCl2 solution

(4) Assuming 100% dissociation NaCl  BaCl2

Sol. Answer (4)

Only solvent particles will move through SPM.
 = iCRT

21. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s)
is(are) [JEE(Advanced) 2013]
(1) G is positive (2) Ssystem is positive
(3) Ssurroundings = 0 (4) H = 0
Sol. Answer (2, 3, 4)
Benzene and naphthalene form an ideal solution. For an ideal solution, H = 0, S system > 0 and
Ssurroundings = 0 because there is no exchange of heat energy between system and surroundings.

22. Mixture(s) showing positive deviation from Raoult's law at 35°C is/are [JEE(Advanced) 2016]
(1) Carbon tetrachloride + methanol (2) Carbon disulphide + acetone
(3) Benzene + toluene (4) Phenol + aniline
Sol. Answer (1, 2)
Mixture Nature of deviation from Raoult's Law
Carbon tetrachloride + methanol +ve
Carbon disulphide + acetone +ve
Benzene + toluene Ideal solution
Phenol + aniline –ve deviation

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Solutions of Assignment Solutions 55
23. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction
of M in solution is shown in the following figure. Here xL and xM represent mole fractions of L and M,
respectively, in the solution. The correct statement(s) applicable to this system is(are)

pL

1 xM 0
[JEE(Advanced) 2017]
(1) Attractive intermolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger
than those between L-M when mixed in solution
(2) The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when xL  0
(3) The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed from xL = 0 to xL = 1
(4) The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when xL  1
Sol. Answer (1, 4)

pL

xM = 1 xM  xM = 0
Pure M xL = 1
Pure L
Point Z represents v.p. of pure liquid L
At xL  1, solution is very dilute, L becomes solvent. Very dilute solution of M in L is nearly ideal and obey
Raoults law (pL = xLpL°)
Also, there is positive deviation indicated by graph above dotted line (expected for ideal solution)
 L-M interaction < L - L & M - M Interactions.

SECTION - C
Linked Comprehension Type Questions
Comprehension-I
The experimental values of colligative properties of many solutes in solution resembles calculated value of
colligative properties.
However in same cases, the experimental value of colligative property differ widely than those obtained by
calculation. Such experimental values of colligative properties are known as Abnormal values of colligative
property. Cause for abnormal values of colligative properties are :
(i) Dissociation of solute : It increases the colligative properties.
e.g. : Dissociation of KCl, NaCl etc. in H2O.
(ii) Association of solute : It decreases the colligative properties
e.g. : Dimerisation of acetic acid in benzene
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56 Solutions Solutions of Assignment

1. If degree of dissociation of an electrolyte A2B3 is 25% in a solvent, then

(1) Normal boiling point = Experimental boiling point
(2) Normal freezing point > Experimental freezing point

1
(3) Normal osmotic pressure = Experimental osmotic pressure
2
1
(4) Normal molecular weight = Experimental molecular weight
4
Sol. Answer (3)
Electrolyte A2B3 is 25% dissociated



 2A3+ + 3B2–
A2B3  (n = 5)

i = 1 + (5 – 1) (0.25) = 1 + (4 × 0.25) = 2
we know that

OBS
CAL = 2

OBS
 CAL =
2

1
or normal osmotic pressure = (observed osmotic pressure)
2

2. 4 different 100 ml solutions are prepared by mixing 1 gram each of NaCl, (NH2)2CO, Na2SO4 and K4[Fe(CN)6]
at temperature T. Correct order of osmotic pressure is

(1) (NH2)2CO solution > NaCl solution > Na2SO4 solution > K4[Fe(CN)6] solution

(2) NaCl solution > Na2SO4 solution > (NH2)2CO solution > K4[Fe(CN)6] solution

(3) K4[Fe(CN)6] solution > Na2SO4 solution > NaCl solution > (NH2)2CO solution

(4) Na2SO4 solution > (NH2)2CO solution > NaCl solution > K4[Fe(CN)6] solution

Sol. Answer (2)

For I – Solution

1
nNaCl = in 100 ml
58.5

1 10
 In 100 ml  MOLES =
58.5

10
1 = × 2 × ST = (0.341) ST
58.5
For II – Solution, UREA

1 ST
= × 10 × 1 × ST =
60 6
 (0.16) ST

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Solutions of Assignment Solutions 57
For III – Solution, Na2SO4

⎛ 1 ⎞ 3  1 ST
10 × ⎜ ⎟ × (3) × ST = = (0.02) ST × 10
⎝ 142 ⎠ 142
 (0.2) ST
For IV – Solution, (K4[Fe(CN)6])

1
10 × × (5) × (S × T) = (0.0135)ST ×10 = (0.1355) ST
368
 Order of osmotic pressure is NaCl soln > Na2SO4 soln > UREA soln > K4[Fe(CN)6] soln

3. One mole I2 (solid) is added in 1 M, 1 litre KI solution. Then

(1) Osmostic pressure of solution increases
(2) Freezing point of solution increases
(3) Relative lowering in vapour pressure decreases
(4) No change in boiling point of solution
Sol. Answer (4)

KI + I2  KI3, KI3 dissociates as K+ & I3–

So, numbers of ions are same. Hence, no change in boiling point of solution.

Comprehension-II
In non-ideal solutions, at one of the intermediate compositions, the total vapour pressure is highest and the
boiling point is lowest. At this point, the composition of the liquid and vapour phase is same. So, if liquid
mixture vapourises at this point and vapours are condensed, the condensate contains same composition as
present in original liquid mixture. It means at this point liquid behaves like a pure liquid and is called an
Azeotropic mixture.

1. Choose the correct statement

(1) Ideal solutions cannot be separated into their components by fractional distillation
(2) For ideal solutions enthalpy of mixing is always greater than zero
(3) Only non-ideal solution showing positive deviation cannot be separated out by fraction distillation
(4) Non-ideal solution showing both positive and negative deviation cannot be separated out by fractional
distillation
Sol. Answer (4)
Non-ideal solution showing +ve and –ve deviation cannot be separated by fractional distillation.

2. A and B forms non-ideal solution showing positive deviation. Boiling point of pure A and B is 350 K and 380
K respectively. The solution will boil at (approximate)
(1) 350 K (2) 380 K (3) > 380 K (4) < 350 K
Sol. Answer (4)
The solution will boil at a temperature less than 350 K, as it forms low boiling point azeotrope.

3. Which of the following cannot form low boiling point azeotrope?

(1) n-heptane & n-hexane (2) Acetone & aniline
(3) Both (1) & (2) (4) CHCl3 & C2H5OH

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58 Solutions Solutions of Assignment

Sol. Answer (3)

For a solution to form low boiling azeotrope, it is necessary that the solution shows positive deviation from
Raoult’s law. N-heptane & n-hexane is nearly an ideal solution while Acetone & Aniline show negative deviation
from Raoult’s law.

Comprehension-III
On mixing two components A and B to form a non-ideal solution, final solution becomes cold.
(F = intermolecular force of attraction).

1. Correct statement about final solution is

(1) Evolution of some heat takes place (2) FAB > FAA and FAB > FBB
(3) Intermolecular H-bonds may be broken out (4) Contraction of volume takes place
Sol. Answer (3)
Fact.

2. If final solution becomes hot, incorrect option

(1) Exothermic dissolution may takes place (2) Vapour pressure increases than expected value
(3) Maximum boiling azeotropes are formed (4) Negative deviations from Raoult’s law
Sol. Answer (2)
Fact.

3. 0.5 M, 100 ml A and 0.2 M, 500 ml B are mixed at 27°C. Vapour pressure of pure A and pure B is 200 mm
Hg and 50 mm Hg respectively at 27°C. Then the ratio of partial pressures of A and B (in vapour phase) after
mixing is
(1) 2 : 1 (2) 1 : 2 (3) 2 : 3 (4) 4 : 1
Sol. Answer (1)

PAo  A
YA 
PAo  A  PBo B

PBo B
YB 
PAo  A  PBo B

PA Y
 A
P B YB

Comprehension-IV
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute
molecules are added to get homogeneous solution. These are called colligative properties. Applications of
colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and
water mixture as anti-freezing liquid in the radiator of automobiles. [IIT-JEE 2008]
A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.
Given : Freezing point depression constant of water (Kfwater) = 1.86 K kg mol–1
Freezing point depression constant of ethanol (Kfethanol) = 2.0 K kg mol–1
Boiling point elevation constant of water (Kbwater) = 0.52 K kg mol–1
Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol–1
Standard freezing point of water = 273 K
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Solutions of Assignment Solutions 59
Standard freezing point of ethanol = 155.7 K
Standard boiling point of water = 373 K
Standard boiling point of ethanol = 351.5 K
Vapour pressure of pure water = 32.8 mm Hg
Vapour pressure of pure ethanol = 40 mm Hg
Molecular weight of water = 18 g mol–1
Molecular weight of ethanol = 46 g mol–1
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be
non-volatile and non-dissociative.
1. The freezing point of the solution M is
(1) 268.7 K (2) 268.5 K (3) 234.2 K (4) 150.9 K
Sol. Answer (4)

0.1 1000
Tf = kfm = 2 × × = 4.83
0.9 46

Tf = T°f – Tf
Tf = 155.7 – 4.83  150.9

2. The vapour pressure of the solution M is

(1) 39.3 mm Hg (2) 36.0 mm Hg (3) 29.5 mm Hg (4) 28.8 mm Hg
Sol. Answer (2)
Water is solute
P0  PS
= Xsolute
P0
40  PS
= 0.1
40

PS = 36 mm of Hg
Here ethanol is considered as non-volalite solute as given in statement of paragraph.

3. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling
point of this solution is
(1) 380.4 K (2) 376.2 K (3) 375.5 K (4) 345.7 K
Sol. Answer (2)
Here, water is solvent and ethanol is solute.
Tb = kb × m

0.1 1000
= 0.52 ×
0.9  18

= 3.2084
Now, B.P. of solution = 373 + 3.2084 = 376.2084

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60 Solutions Solutions of Assignment

SECTION - D

Assertion-Reason Type Questions

1. STATEMENT-1 : On cooling a mixture of ideal gases, an ideal solution can be obtained.

and
STATEMENT-2 : Ideal solution do not form azeotropes.
Sol. Answer (4)
It is the non-ideal solution which are having (H)mix  0, (V)mix  0 which form azeotropes and ideal solutions
can not be obtained by cooling gases as the gases have different TC.

2. STATEMENT-1 : Relative lowering of vapour pressure is equal to mole fraction of the solute.
and
STATEMENT-2 : Relative lowering of vapour pressure is a colligative property.
Sol. Answer (2)

P  PS
is relative lowering of vapour pressure and is equal to mole fraction of solute because pressure of
P
solution of non-volatile solute is proportional to mole fraction of solvent.

P  PS
is a colligative property
P

 Both are correct but (2) is not correct explanation.

3. STATEMENT-1 : When HgI2 is added to the aqueous solution of KI, the freezing point is raised.
and
STATEMENT-2 : Freezing point generally increases by adding non volatile solute in solvent.
Sol. Answer (3)
When HgI2 is added it reacts with KI
HgI2 + 2KI  K2(HgI4)
 Solute particles decreases. Tf is depression in FP and is decreased by adding non-volatile solute.

4. STATEMENT-1 : At low concentration benzene and toluene forms ideal solution.

and
STATEMENT-2 : Components with structural similarities form ideal solution.
Sol. Answer (1)
Benzene and Toluene form an ideal solution because their molecules are identical in shape, size and nature.

5. STATEMENT-1 : Molality and mole fraction are temperature independent quantity.

and
STATEMENT-2 : Molality and mole fraction are unit less quantity.
Sol. Answer (3)
Molality involves weight and mole fraction involves moles that’s why they are temperature independent quantity.
Molality have units as ‘molal’ but mole fraction does not have units.

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Solutions of Assignment Solutions 61
6. STATEMENT-1 : 0.1 M solution of Na2SO4 has greater osmotic pressure than 0.1 M solution of urea at same
temperature.
and
STATEMENT-2 : The value of van’t Hoff factor for Na2SO4 is less than urea.
Sol. Answer (3)
For Na2SO4 i=3
and 1 = (0.1) (ST) × 3
and for UREA i=1
and 2 = (0.1) (S) T
 1 > 2
i for Na2SO4 = 3 ; and for UREA, i=1
 It can be concluded that osmotic pressure for Na2SO4 solution is more as compared to ‘UREA’ solution.
 Statement (1) is correct but statement (2) is wrong.

M
7. STATEMENT-1 : The equivalent mass of Mohr’s salt is in non-redox reaction [If M is molecular mass of
4
Mohr’s salt].
and
STATEMENT-2 : The normality of Mohr’s salt is higher than molarity for same amount and volume at constant
temperature.
Sol. Answer (2)

M
Mohr’s salt is FeSO4.(NH4)2 SO4.6H2O. So, Equivalent wt. = .
4
Normality = Molarity × n-factor.

8. STATEMENT-1 : One molar solution is always more concentrated than one molal solution.
and
STATEMENT-2 : The amount of solvent in 1 M and 1 m aqueous solution is not equal.
Sol. Answer (4)
Fact.

9. STATEMENT-1 : Iodine is more soluble in CCl4 than in water.

and
STATEMENT-2 : Non-polar solutes are more soluble in non-polar solvents
Sol. Answer (1)
Fact.

10. STATEMENT-1 : Henry’s law and Raoult’s law are not independent, i.e., one can be derived from the other.
and
STATEMENT-2 : The partial pressure is directly proportional to the mole fraction of concerned species for ideal
solutions.
Sol. Answer (2)
Raoult’s law can be derived from Henry’s law.

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62 Solutions Solutions of Assignment

SECTION - E

Matrix-Match Type Questions

1. Match the following

Column-I Column-II
(A) Acetone + Carbon disulphide (p) Vapour pressure measurement
(B) Acetone + Aniline (q) Osmotic pressure
(C) Used in calculation of molar (r) Maximum boiling azeotrope
mass of polymer
(D) Ostwald-Walker’s method (s) Minimum boiling azeotrope
Sol. Answer A(s), B(r), C(q), D(p)
(A) Acetone and CS-2 forms minimum boiling point azeotrope.
(B) CH3COCH3 and C6H5NH2 forms maximum B.P. azeotrope.
(C) Barkley’s and Hartley method is used to determine osmotic pressure.
(D) Ostwald –Walker method are used to measure the vapour pressure.

2. Match the following

Column-I Column-II
Solute (equimolar) (assuming  = 100%)  (osmotic pressure) ratio
(A) NaCl, MgCl2, K2SO4 (p) 1 : 2 : 3
(B) Al2(SO4)3, Na3PO4, K4[Fe(CN)6] (q) 1 : 1 : 1
(C) Glucose, Rock salt, Glauber’s salt (r) 1 : 0.8 : 1
(D) Baryta water, lime water, washing soda (s) 2 : 3 : 3
Sol. Answer A(s), B(r), C(p), D(q)
(A) NaCl  Na+ + Cl–
i=1+=2
MgCl2  Mg2+ + 2Cl–
i = 1 + 2 = 3
K2SO4  2K+ + SO42–
i = 1 + 2 = 3
 Ratio is 2 : 3 : 3
(B) Al2(SO4)3  2AI3++3SO42–
For strong electrolyte  = 1
i=1+4=5
Na3PO4  3Na++ PO43–
i=1+3=4
K4[Fe(CN)6] = 1 + 4 = 5
 Ratio is 5 : 4 : 5
 1 : 0.8 : 1

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Solutions of Assignment Solutions 63
(C) C6H12O6 i=1
Rock Salt i=2
Glauber’s Salt (Na2SO4) i = 3
Ratio is 1: 2 : 3
(D) Baryta water Ba(OH)2 ; i = 3
lime water Ca(OH)2 ; i = 3
Washing Soda Na2CO3 ; i = 3
Ratio is 1 : 1 : 1

3. Match the following

Column-I Column-II
(A) 100 ml 1.2% (NH2)2CO solution (wt./vol) at (p) Isotonic with 100 ml 1M K3[Fe(CN)6] at
T temperature T temperature
(B) 250 ml 2 M NaNO3 solution at (q) Isotonic with 100 ml 6.84% (wt./vol) C12H22O11 at
T temperature T temperature
(C) 5.845% (by wt.) NaCl aqueous solution (r) Boiling point = 100.052°C
(D) 1.8% (by wt.) C6H12O6 aqueous solution (s) Boiling point = 101.1°C
Sol. Answer A(q), B(p), C(s), D(r)
(Column – I)
(A) 1.2% UREA is at temp. T
M (UREA) = 60
1.2
 UREA = in 100 ml
60
1.2 12 1
 In 1000 ml ‘C’ = × 10 = =
60 60 5
1 ST
 = 1 × 0.0821 × T =
5 5
(B)  = 2 × 2 × ST = 4ST
(0.52)  5.845  1000  2
(C) Tb = = 1.10°C
58.5  94.155
(0.52)  1.8  1000
(D) Tb = = 0.05°C
180  98.2
(Column – II)
(p)  = 4 × 1× ST = (4ST)
1 6.84  10
(q)  = × 0.0821 × T
342
 (0.2 ST)
Therefore it can be seen

⎛ ST ⎞
(A) and (q) have same osmotic pressure i.e. ⎜ ⎟
⎝ 5 ⎠
(B) and (p) are isotonic solution and have same osmotic pressure
(C)  (s) as Tb = 1.04°C
(D)  (r) as Tb = 0.05°C

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64 Solutions Solutions of Assignment

4. Match the following

Column-I Column-II
(A) Benzene + Carbon tetrachloride (p) Form maximum boiling azeotrope

(B) Chlorobenzene + Water (q) PTotal  poA  pBo

(C) Ethanol + Water (r) Form minimum boiling azeotrope

(D) Acetone + Chloroform (s) Does not form azeotrope
Sol. Answer A(r), B(q, s), C(r), D(p)

5. Match the following

Column-I Column-II
(Solute, degree of ionisation) (van't Hoff factor)
(A) K2SO4 ;  = 0.85 (p) 3.7
(B) SnCl2 ;  = 0.7 (q) 4.0
(C) Al(NO3)3 ;  = 0.9 (r) 2.7
(D) Fe2(SO4)3 ;  = 0.75 (s) 2.4
Sol. Answer A(r), B(s), C(p), D(q)

SECTION - F

Integer Answer Type Questions

1. For [CrCl3.xNH3], elevation in boiling point of one molal solution is triple of one molal aqueous solution of urea.
Assuming 100% ionisation of complex molecule and coordination number as six, calculate the value of x.
Sol. Answer (5)
Tb (complex) = 3 × Tb (area)

Thus, complex should furnish three ions. Therefore, complex is CrCl.xNH3  Cl2  CrCl.xNH3 
2
 2Cl
Also, co-ordination number of Cr is six
Thus, x = 5

2. An aqueous solution of an acid is so weak that it can be assumed to be practically unionised, boiled at 100.4°C
25 ml of this solution was neutralised by 38.5 ml of 1 N solution of NaOH. Calculate basicity of the acid if
kb(H2O)=0.52 k mol–1kg. Assume molality is equal to molarity.
Sol. Answer (2)

Tb  k b  m

0.4 = 0.52 × m  m = 0.77

25 × 0.77 × ‘n’ = 38.5 × 1  n = 2

3. The vapour pressure of two pure liquids A and B that forms an ideal solution are 300 and 800 torr respectively
at temperature T. A mixture of the vapours of A and B (for which the mole fraction of A is 0.25) is slowly
compressed at temperature T. The vapour pressure of this condensate on 100% condensation of vapoure is
measured to be 'P'. What is the value of 684 - P?

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Solutions of Assignment Solutions 65
Sol. Answer (9)

X 'A poA X A X A X 'A pBo 0.25 800 8

 ⇒     
X 'B pBo XB XB X 'B poA 0.75 300 9

 XB = 0.53

 P = 300 × 0.47 + 800 × 0.53 = 565 torr

So, 574 – P = 9

4. A compound X undergoes tetramerisation in a given organic solvent. The van’t Hoff factor i is calculated as
0.05y. Find y. (Assuming 100% association)

Sol. Answer (5)

4X  (X)4


i  (1   ) 
n

3
  1, n  4 ⇒ i  1   0.25
4

0.25 = 0.05 y
 y=5

5. If K3[Fe(CN)6] gets ionized completely in a solution, number of particles in the solution from 1 molecule solute
is________.

Sol. Answer (4)

K3[Fe(CN)6]  3K+ + [Fe(CN)6)]3–

6. MX2 dissociates into M2+ and X– ions in an aqueous solution, with a degree of dissociation (a) of 0.5. The ratio
of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing
point in the absence of ionic dissociation is [JEE(Advanced)-2014]
Sol. Answer (2)

 = 0.5

MX2 M+2 + 2X–

(1 – )  2

1      2
i
1

i = 1 + 2
i = 1 + 2 × 0.5 = 2

7. If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex (which behaves
as a strong electrolyte) is –0.0558°C, the number of chloride(s) in the coordination sphere of the complex is
[Kf of water = 1.86 K kg mol–1] [JEE(Advanced)-2015]

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66 Solutions Solutions of Assignment

Sol. Answer (1)

Given Tf = 0.0558°C
as we know, Tf = i × Kf × m
 0.0558 = i × 1.86 × 0.01

i3
Therefore the complex will be [Co(NH3)5Cl]Cl2
Hence number of chloride in co-ordination sphere is 1.

8. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution
of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole
fractions xA and xB, respectively, has vapour pressure of 22.5 Torr. The value of xA/xB in the new solution is ___.
(given that the vapour pressure of pure liquid A is 20 Torr at temperature T) [JEE(Advanced)-2018]
Sol. Answer (19.00)
P°A = 20 Torr
1
For equimolar binary solution : xA = xB =
2

PA  PB
 = 45
2
P°B = 70 Torr
If mole fractions are xA & xB
P°B + (P°A – P°B)xA = 22.5
70 + (20 – 70)xA = 22.5

47.5 2.5
xA = and xB =
50 50

x A 47.5

xB 2.5 = 19.00

9. The plot given below shows P – T curves (where P is the pressure and T is the temperature) for two solvents
X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

12
3 4
Pressure (mmHg)

760
1. solvent X
2. solution of NaCl in solvent X
3. solvent Y
4. solution of NaCl in solvent Y
360
362

367
368

Temperature (K)
On addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the
elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization
in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is _____.
[JEE(Advanced)-2018]

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Solutions of Assignment Solutions 67
Sol. Answer (0.05)
When NaCl as solute is used
For solvent X For solvent Y
2 = 2Kb m 1 = 2 × Kb m

Kb
 K b = 2
When solute S is used then molality in both solvent is equal.
For solvent X For solvent Y

 0.7
i = 1– i = 1– = 0.65
2 2
⎛ ⎞
Tb = ⎜ 1– ⎟ K bm Tb = (0.65)Kbm
⎝ 2⎠

⎛ ⎞
⎜ 1– ⎟  2
Tb ⎝ 2⎠
3= 
Tb 0.65

 3
1– =  0.65
2 2
 3
= 1–  0.65
2 2
 = 0.05

SECTION - G
Multiple True-False Type Questions

1. STATEMENT-1 : Solubility of a gas in a liquid solution as per Henry law is a function of the partial pressure
of the gas at constant temperature.
STATEMENT-2 : Mole fraction of the gas in a solution as per Henry law is proportional to the partial pressure
of the gas at constant temperature.
STATEMENT-3 : As per Henry law higher the value of kH at a given partial pressure and temperature lower is
the solubility of the gas in the liquid.
(1) T T T (2) FTT (3) FTF (4) TFF
Sol. Answer (1)

2. STATEMENT-1 : For solution of volatile liquids, the partial vapour pressure of each component in the solution
is directly proportional to its mole fraction.
STATEMENT-2 : Always there will be lowering of vapour pressure on addition of non-volatile solute to a solvent.
STATEMENT-3 : If there is dissociation of non-volatile solute then the V.P. of solution increases.
(1) T F T (2) FFT (3) TTF (4) FTF
Sol. Answer (3)

p0  p in

p 0
in  N
i < 1 or i > 1 (doesn’t matter)

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68 Solutions Solutions of Assignment

3. STATEMENT-1 : Effect of adding a non-volatile solute to a solvent is to increase its freezing point.
STATEMENT-2 : Molality is a dimensionless quantity.
STATEMENT-3 : The hard shell of an egg was dissolved in HCl solution, and then egg was placed in
concentrated solution of NaCl. Then egg will shrink.
(1) T F T (2) FFT (3) TTF (4) FTF
Sol. Answer (2)
Tf = (freezing point)solvent – (freezing point)solution
4. STATEMENT-1 : At definite temperature, the solubility of a solute is fixed.

STATEMENT-2 : When azeotropic mixture is distilled, its composition remains same.

STATEMENT-3 : Vapour pressure is a colligative property.
(1) F F T (2) TTF (3) F TF (4) TTT
Sol. Answer (2)
Vapour pressure is not a colligative property, rather relative lowering of vapour pressure is a colligative property.

5. STATEMENT-1 : 1 M H2SO4 is also 1 N.

STATEMENT-2 : Water glycol mixture is used in car radiators in winter because its freezing point is less than
0°C.
STATEMENT-3 : A saturated solution will remain saturated at all temperatures.
(1) T F T (2) FTT (3) FTF (4) T FF
Sol. Answer (3)
Normality = molarity × n-factor
A saturated solution is saturated at a given temperature.

SECTION - H
Aakash Challengers Questions
1. An aqueous solution of H2SO4 has density 1.84 g/ml. Solution contains 98% H2SO4 by mass. Calculate
(i) Molarity of solution
(ii) Molar volume of solution
(iii) Relative lowering of vapour pressure w.r.t. water, assuming H2SO4 as non-electrolyte at this high
concentration.
98 1000
Sol. (i) Molarity, M    1.84  18.4 M
98 100
1
(ii) nH2 SO4  1, nwater 
9
⎛ 1⎞
1 ⎜ ⎟
9
Molar mass of solution = 98   18  ⎝ ⎠  90 gram
⎛ 1⎞ ⎛ 1⎞
⎜1  ⎟ ⎜1  ⎟
⎝ 9⎠ ⎝ 9⎠

90
Molar volume of solution =  48.916 ml/mole
1.84

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Solutions of Assignment Solutions 69

P  P 1
(iii)   0.9
P ⎛ 1⎞
⎜ 1  ⎟
⎝ 9⎠

2. o
Benzene and toluene forms nearly an ideal solution. At 300 K, Ptoluene o
= 32.06 mm and Pbenzene  103.01 mm
(of Hg)

(i) A liquid mixture is composed of 3 mole of toluene and 2 mole of benzene. If the pressure over the mixture
at 300 K is reduced, at what pressure does the first vapour form?

(ii) What is the composition of the first trace of vapour formed?

(iii) If the pressure is reduced further, at what pressure does the last trace of liquid dissappear?

(iv) What is the composition of last trace of liquid?

3 2
Sol. (i) P  32.06   103.01  60.44 mm Hg
5 5

1 yt 1 yt
(ii)   ⇒ y toluene  0.3181
60.44 32.06 103.01

3 2
3 1
(iii) Assuming liquid containing xtoluene = distilled from vapour,  5  5 ⇒ P  44.25 mm Hg
5 P 32.06 103.01

(iv) 44.25 = xt × 32.06 + (1 – xt) × 103.01

 xt = 0.8281

3. When 45 g of an unknown compound was dissolved in 500 g of water, the solution has freezing point of
–0.93 °C

(i) What is the molecular weight of compound? (kf = 1.86)

(ii) If empirical formula is CH2O, what is the molecular formula of compound?

45 1000
Sol. (i) 0.93  1.86   ⇒ M  180
M 500

180
(ii) n   6  molecular formula = C6H12O6
(12  2  16)

4. A complex is represented as CoCl 3 .xNH 3 . Freezing point of its 0.1 molal solution is –0.558°C
(kf = 1.86). Assuming 100% dissociation and coordination number of Co(III) is six, then what is the complex?
Sol. 0.558 = i × 1.86 × 0.1  i = 3
 complex ionizes to form three ions
Since, C.N. = 6  x = 5
CoCl3.5NH3  [Co(NH3)Cl]2+ + 2Cl–
 Complex is [Co(NH3)5Cl]Cl2

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70 Solutions Solutions of Assignment

5. A certain mass of substance in 100 g of benzene lowers the freezing point by 1.28°C and in
100 g of water lowers the freezing point by 1.395°C separately. If the substance has normal molecular weight
in benzene and completely dissociated in water, calculate number of moles of ions formed by its 1 mole
dissociation in water. ( K f(water) = 1.86, K f(benzene) = 5.00)

Sol. In water, 1.28 = 1.86 × m

1.28
In benzene, 1.395 = i × 5.00 × ⇒i3
1.86

Since,  = 100%, number of particles dissociated from 1 mole = 3.

6. The freezing point of nitrobenzene is 3°C. When 1.2 g chloroform (mol. wt. = 120) is dissolved in 100 g of
nitrobenzene, freezing point will be 2.3°C. When 0.6 g of acetic acid is dissolved in 100 g of nitrobenzene,
freezing point of solution is 2.64°C. If the formula of acetic acid is (CH2O)n, find the value of n.

1.2 1000
Sol. 3 – 2.3 = K f   ⇒ Kf  7
120 100

0.6 1000
3 – 2.64 = 7   ⇒ M  117
M 100

117
n 4
30

7. How much ice will be separated out if a solution containing 25 gram of ethylene glycol [C2H4(OH)2] in 100 gram
of water is cooled to –10°C? (K f( water )  1.86) ?

25 1000
Sol. 0  ( 10)  1.86   ⇒ W  75 gram
62 W

Amount of ice separated = 100 – 75 = 25 g

8. A 0.075 molar solution of monobasic acid has a freezing point of –0.18°C. Calculate Ka for the acid, (kf = 1.86)
Sol. 6.75 × 10–3

9. The freezing point of an aqueous solution of KCN containing 0.1892 mole/kg H2O was –0.704°C. On adding
0.095 mole of Hg(CN)2, the freezing point of solution was –0.53°C. Assuming that complex is formed according
to the reaction

Hg(CN)2  xCN–  Hg(CN)xx  2

and also Hg(CN)2 is limiting reagent, find x.

Tf 0.704 (KCN  K   CN– )

Sol. k f  
i  m 2  0.1892 100% dissociation

= 1.86

Hg(CN)2 + xCN–  Hg(CN)xx  2

0.095 0.1892 0
0 (0.1892 – x × 0.095) 0.095

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Solutions of Assignment Solutions 71
Total molality after adding Hg(CN)2,
= molality of K+
+
molality of CN–
+

molality of Hg(CN)xx  2 remaining

= 0.1892 + 0.1892 – 0.095x + 0.095

= 0.4734 – 0.095x

Tf 0.53
Kf  ⇒ 1.86 
m 0.4734  0.095x

 x2

10. 0.01 molal aqueous solution of K3[Fe(CN)6] freezes at –0.062°C. Calculate percentage dissociation. (kf = 1.86)
Sol. Tf(normal) = Kf × m = 0.01 × 1.86
= 0.0186

( Tf )obs. 0.062

i 
( Tf )normal 0.0186


  3
 3K  [(Fe(CN)6 )]
K 3 [Fe(CN)6 ] 
(1 ) 3 

1    3   0.062
i 
1 0.0186

  = 0.78

11. Vapour pressure of mixture of liquid A and liquid B at 70°C is given by PT = 180B + 90(in mm), where B is
the mole fraction of B, in the liquid mixture. Calculate
(a) Vapour pressure of pure A and pure B
(b) Vapour pressure of mixture of A and B by mixing 4 g A and 12 g B. (If molar mass of A and B are 2 g
and 3 g respectively)
(c) From (b) ratio of moles of A and B in vapour at 70°C
Sol. Total pressure as a function of XB is given by
PT = 180 XB + 90
(a) When XB = 0, XA = 1

PT = (180 × 0) + 90 = 90 mm Hg = PAo

 PAo = 90 mm Hg

and XB = 1; XA = 0

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72 Solutions Solutions of Assignment

 PT = (180 × 1) + 90 = 270 mm Hg = PBo

Hence the vapour pressures of pure A & pure B

PAo = 90 mm Hg

PBo = 270 mm Hg

(b) wA = 4 gm; MA = 2
wB = 12 gm; MB = 3
 Number of moles

wA 4 wB 12
nA = M = = 2; nB = M = =4
A 2 B 13
nA = 2; nB = 4

PT = PAo XA + PBo XB

90  2 270  4
 PT = +
24 24

270  2
 PT = 30 + = 30 + 180
3
 Total vapour pressure is 210 mm Hg = PT

yA
(c) We have to calculate y
B

PAo XA PBo XB
yA = and yB =
PT PT

yA PAo X A
 =
yB PBo XB

yA 90 2 n A nT nA 1
 =  =   n =
yB 270 4 nT nB B 6

12. Two elements x and y form two types of molecules like xy3 and xy. 1 gm of xy3 lowers the freezing point by
1°C when dissolved in 100 gram solvent whereas for the same lowering in freezing point for same amount of
solvent 0.5 gram xy is required. Calculate atomic mass of x and y. (Given : Kf for solvent is 5 K molal–1).
Sol. Part-I
For ‘xy3’
wsolute = 1 g; (Tf)1 = 1ºC
wsolvent =100 g; Kf = 5
Tf = Kf × m

5  1 1000
 1= M1  100

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Solutions of Assignment Solutions 73

5  1000
 M1 = = 50
100

50 = x + 3y …(i)
For ‘xy’
wsolute = 0.5 g; wsolvent = 100 g
(Tf)2 = 1ºC; Kf = 5

5  0.5  1000
1= M2  100

5  0.5  1000
 M2 = = 25
100

25 = x + y …(ii)
Solving (i) & (ii), we get
2y = 25, y = 12.5 & x = 12.5

13. 17.6 gram of unknown solute is dissolved in 100 gram of a solvent (Kb = 2 K molality–1) to prepare a solution.
Boiling point of pure solvent is 225°C where as boiling point of this solution is 229°C. Predict molecular formula
of solute if it contains 54.54% C and 9.09% H (by weight).
Sol. Given :
wsolute = 17.6 g
wsolvent = 100 g
Kb = 2
Tb (in problem) = 229 – 225 = 4ºC
According to the relation, elevation in boiling point
Tb = Kb × m

17.6  1000
4 = (2)
M  100

 M (Molecular mass)

2  17.6  1000
 M= = 88
100  4

54.54% 4.5
C= = =2
12 2.27

9.09% 9.09
H= = =4
1 2.27

36.37% 2.27
O= = =1
16 2.27
Therefore, empirical formula is C2H4O
 Molecular formula is C4H8O2.

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74 Solutions Solutions of Assignment

M
14. What weight of 60% pure NaOH is required to neutralise 100 ml H SO solution?
10 2 4

Sol. Let the total weight is ‘x’ gm

x  60
60% NaOH i.e., = wNaOH
100

60x
Equivalents of NaOH =
100  40

According to law of equivalents:

eq (NaOH) = eq (H2SO4)

60 x 100  1
 = × 2 ×10–3
100  40 10

100  2  10 –3  100  40
 x=
60  10  1

 wNaOH = 1.33 g

15. A 0.5% (by weight) solution of A2B in solvent C was found to freeze at –3.25ºC. Calculate the degree of
dissociation of A2B in solvent C into A–2 and B+. (Given freezing point of pure C is –3°C, molar weight of A2B
is 60 and Kf of C is 2 K–1 molality–1).
Sol. wsolvent = 100 g – 0.5 g = 99.5 g
wsolute = 0.5 g
Tf = 0.25ºC, Kf = 2

MA2B = 60

Tf = i × Kf × m

i  2  0.5 1000
(0.25) = ×
60 99.5

(0.25  60)  99.5

i= = 1.5
2  0.5  1000

We know
i = 1 + (n – 1) 
1.5 = 1 + (2 – 1)
 = 0.5
 It is 50% dissociated.

16. If boiling point of an aqueous solution is 100.1°C. What is its freezing point. (Given : latent heat of fusion and
vapourisation of water are 80 cal gram–1 and 540 cal gram–1 respectively)?
Sol. From the latent heat of fusion,
Kb = 0.52 k/m & Kf = 1.86 k/m
Tb = Kb × m

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Solutions of Assignment Solutions 75

0.1
m=
0.52

Tf = Kf × m

0.1
= 1.86 × = 0.357
0.52

Hence, Freezing point of solution = 0 – 0.357 = – 0.357ºC

17. Osmotic pressure of an aqueous solution at 27°C is found to be 1900 mm Hg. What will be the freezing point
of solution (Assuming, molality = molarity × 1.5)? (Kf = 1.86)

1900
Sol.  (osmotic pressure) = atm
760

T = 27ºC + 273 = 300 K

 According to law of Vant Hoff
=C×S×T

1900
 C= = 0.101 M
760  300  0.0821

According to problem
Molality = (1.5 × 1.101) M = 0.152
We know that
Tf = Kf × m
Tf = 1.86 × 0.152 = 0.283ºC
Hence freezing point = 0 – (+ 0.283) = – 0.283ºC

18. In Ostwald and Walker’s apparatus, dry air is passed through a solution containing 20 gram of an organic
non-volatile solute in 250 ml of water. Then the air was passed through pure water and then through a U-tube
containing anhydrous CaCl2. The mass lost in solution is 26 gram and the mass gained in the U-tube is
26.48 gram. Calculate the molecular mass of organic solute.
Sol. From Ostwald & Walker’s process,

Loss in mass of solvent P0 – P w M

= = 
Gain in mass of CaCl2 P0 m W

26.48 – 26 20 18
= 
26 m 250

0.48 20 18
= 
26 m 250

20  18  26
m = 250  0.48 = 78 g

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76 Solutions Solutions of Assignment

19. The vapour pressure of pure benzene is 639.70 mm of Hg and the vapour pressure of solution of a solute in
benzene at the same temperature is 631.9 mm of Hg. Calculate the molality of the solution.
Sol. P = P0x1
631.9 = 639.7x1
x1 = 0.9878
x2 = 1 – x1
x2 = 1 – 0.9878 = 0.0122

0.0122
Molality =  1000  0.158
0.9878  78

20. 180 ml of pure water at 4°C is saturated with NH3 gas, yielding a solution of density 0.8 gram/ml and containing
NH3 (40% by weight). Calculate
(a) Volume of NH3 solution.
(b) Volume of NH3 present in saturated solution.
(c) Volume of NH3 gas at 15°C and 950 mm of Hg in saturated solution.
Sol. 375 ml, 195 ml, 133.3 ml

  

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