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# Daily Practice Problems 5 Class 12 Notes | EduRev

## Class 12 : Daily Practice Problems 5 Class 12 Notes | EduRev

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9217610408,8699438881,8699438882
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G od has never given dreams
SOLUTIONS
DAILY  PRACTICE  PROBLEMS - 5
NON-MEDICAL ENTRANCE PREVIOUS YEAR QUESTIONS
1. The henry’s law constant for the solubility of N
2
gas in water at 298 K is 1.0 x 10
5
atm. The mole fraction of N
2
in air
is 0.8. The number of moles of N
2
from air dissolved in 10 moles of water at 298 K and 5 atm pressure is(IIT 2009)(a) -4
4.0 x10(b) -5
4.0 x10(c) -4
5.0 x10(d) -6
4.0 x10
2. The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its
aqueous solution at 27
0
C is 25 g mol
—1
. Therefore, its ionization percentage in this solution is(J & K 2009)(a) 75(b) 60(c) 80(d) 70
3. The freezing point of water is depressed by 0.37
0
C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal
solution of urea is depressed by(WBJEE 2008)(a) 0.37
0
C(b) 0.74
0
C(c) 0.185
0
C(d) 0
0
C
4. 0.004 M Na
2
SO
4
is isotonic with 0.01 M glucose. Degree of dissociation of Na
2
SO
4
is(IIT 2004)(a) 75%(b) 50%(c) 25%(d) 85%
5. The freezing point (0
0
C) of a solution containing 0.1 g of K
3
[Fe(CN) 6
] (Mol. wt. 329) in 100 g of water (K
f
= 1.86 K
kg mol
—1
) is(IIT 2011)(a) -2
-2.3 x10
(b) -2
-5.7 x10
(c) -3
-5.7 x10
(d) -2
-1.2 x10
6. A 0.2 molal aqueous solution of a weak acid(HX) is 20% ionised. The freezing point of the solution is (IIT 1995)(a) —0.45
0
C(b) 90
0
C(c) —0.31
0
C(d) —0.53
0
C
7. The molecular weight of benzoic acid in benzene as determined by depression in freezing point method
corresponds to    (IIT 1996)(a) ionization of benzoic acid(b) dimerisation of benzoic acid
(c) trimerization of benzoic acid(d) solvation of benzoic acid
8. The van’t Hoff factor of 0.1 M Ba(NO
3
) 2
solution is 2.74. The degree of dissociation kilogram of water, is(IIT 1999)(a) 91.3%(b) 87%(c) 100%(d) 74%
9. The vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985
N/m
2
. The vapour pressure of water is 3000 N/m
2
. The molecular mass of the solute is(IIT 1993)(a) 60(b) 120(c) 180(d) 380
10. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to (IIT 1983)(a) concentraton of solute in g/L(b) concentration of solute in g/100 mL
(c) mole fraction of solute(d) mole fraction of solvent
Page 2

DPP : 05 / Solutions
SCF-07, 1
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floor,Sector -15, Panchkula
9217610408,8699438881,8699438882
Page No.1
without ability to achieve it.
G od has never given dreams
SOLUTIONS
DAILY  PRACTICE  PROBLEMS - 5
NON-MEDICAL ENTRANCE PREVIOUS YEAR QUESTIONS
1. The henry’s law constant for the solubility of N
2
gas in water at 298 K is 1.0 x 10
5
atm. The mole fraction of N
2
in air
is 0.8. The number of moles of N
2
from air dissolved in 10 moles of water at 298 K and 5 atm pressure is(IIT 2009)(a) -4
4.0 x10(b) -5
4.0 x10(c) -4
5.0 x10(d) -6
4.0 x10
2. The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its
aqueous solution at 27
0
C is 25 g mol
—1
. Therefore, its ionization percentage in this solution is(J & K 2009)(a) 75(b) 60(c) 80(d) 70
3. The freezing point of water is depressed by 0.37
0
C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal
solution of urea is depressed by(WBJEE 2008)(a) 0.37
0
C(b) 0.74
0
C(c) 0.185
0
C(d) 0
0
C
4. 0.004 M Na
2
SO
4
is isotonic with 0.01 M glucose. Degree of dissociation of Na
2
SO
4
is(IIT 2004)(a) 75%(b) 50%(c) 25%(d) 85%
5. The freezing point (0
0
C) of a solution containing 0.1 g of K
3
[Fe(CN) 6
] (Mol. wt. 329) in 100 g of water (K
f
= 1.86 K
kg mol
—1
) is(IIT 2011)(a) -2
-2.3 x10
(b) -2
-5.7 x10
(c) -3
-5.7 x10
(d) -2
-1.2 x10
6. A 0.2 molal aqueous solution of a weak acid(HX) is 20% ionised. The freezing point of the solution is (IIT 1995)(a) —0.45
0
C(b) 90
0
C(c) —0.31
0
C(d) —0.53
0
C
7. The molecular weight of benzoic acid in benzene as determined by depression in freezing point method
corresponds to    (IIT 1996)(a) ionization of benzoic acid(b) dimerisation of benzoic acid
(c) trimerization of benzoic acid(d) solvation of benzoic acid
8. The van’t Hoff factor of 0.1 M Ba(NO
3
) 2
solution is 2.74. The degree of dissociation kilogram of water, is(IIT 1999)(a) 91.3%(b) 87%(c) 100%(d) 74%
9. The vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985
N/m
2
. The vapour pressure of water is 3000 N/m
2
. The molecular mass of the solute is(IIT 1993)(a) 60(b) 120(c) 180(d) 380
10. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to (IIT 1983)(a) concentraton of solute in g/L(b) concentration of solute in g/100 mL
(c) mole fraction of solute(d) mole fraction of solvent
DPP : 05 / Solutions
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9217610408,8699438881,8699438882
Page No.2
without ability to achieve it.
G od has never given dreams
11. Calculate the molal depression constant of a solvent which has freezing point 16.6
0
C and latent heat of
fusion 180.75 J g
—1
[JEE (Orissa) 2005]
(a) 2.68(b) 3.86(c) 4.68(d) 2.86
12. The elevation in boiling point for 13.44g of CuCl
2
dissolved in 1kg of water as solvent will be      (IIT 2005)(a) 0.05(b) 0.10(c) 0.16(d) 0.20
13. The amount of urea to be dissolved in 500 cc of water (K
f
= 1.86) to produce a depression of 0.186
0
C in the freezing
point is          [UGET (Manipal) 2006]
(a) 9 g(b) 6 g(c) 3 g(d) 0.3 g
14. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of
propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same
temperature will be(AIEEE 2007)(a) 360(b) 350(c) 300(d) 700
15. X is a non-volatile solute and Y is volatile solvent. The following vapour pressures are obtained by dissolving X and Y .
X mol L
—1
Y/mm Hg
0.1      P
1
0.25      P
2
0.01      P
3
The correct order of  vapour pressure is [EAMCET (Engg) 2010]
(a) P
1
< P
2
< P
3
(b) P
3
< P
2
< P
1
(c) P
3
< P
1
< P
2
(d) P
2
< P
1
< P
3
16. 18 g glucose glucose (C
6
H
12
O
6
) is added to 178.2g of water. The vapour pressure of water for this aqueous
solution at 100
0
C is(AIEEE 2006)(a) 759 torr(b) 7.60 torr(c) 76 torr(d) 752.4 torr
17. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol
—1
) in the same
solvent. If the denisities of both the solutions are assumed to be equal to 1 g cm
—3
, molar mass of the substance
will be(AIEEE 2007)(a) 210.0 g mol
—1
(b) 90.0 g mol
—1
(c) 115.0 g mol
—1
(d) 105.0g mol
—1
18. At 80
0
C, the vapour pressure of pure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg.
If a mixture of solution A and B boils at 80
0
C and 1 atm pressure, the amount of A in the mixture is(1atm = 760 mm
of Hg)(AIEEE 2008)(a) 50 mol %(b) 52 mol %(c) 34 mol %(d) 48 mol %
19. K
f
for water is 1.86 K kg mol
—1
. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol
(C
2
H
6
O
2
) must you add to get the freezing point of the solution lowered to —2.8
0
C ?(AIEEE 2012)(a) 27 g(b) 72 g(c) 93 g(d) 39 g
Page 3

DPP : 05 / Solutions
SCF-07, 1
st
floor,Sector -15, Panchkula
9217610408,8699438881,8699438882
Page No.1
without ability to achieve it.
G od has never given dreams
SOLUTIONS
DAILY  PRACTICE  PROBLEMS - 5
NON-MEDICAL ENTRANCE PREVIOUS YEAR QUESTIONS
1. The henry’s law constant for the solubility of N
2
gas in water at 298 K is 1.0 x 10
5
atm. The mole fraction of N
2
in air
is 0.8. The number of moles of N
2
from air dissolved in 10 moles of water at 298 K and 5 atm pressure is(IIT 2009)(a) -4
4.0 x10(b) -5
4.0 x10(c) -4
5.0 x10(d) -6
4.0 x10
2. The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its
aqueous solution at 27
0
C is 25 g mol
—1
. Therefore, its ionization percentage in this solution is(J & K 2009)(a) 75(b) 60(c) 80(d) 70
3. The freezing point of water is depressed by 0.37
0
C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal
solution of urea is depressed by(WBJEE 2008)(a) 0.37
0
C(b) 0.74
0
C(c) 0.185
0
C(d) 0
0
C
4. 0.004 M Na
2
SO
4
is isotonic with 0.01 M glucose. Degree of dissociation of Na
2
SO
4
is(IIT 2004)(a) 75%(b) 50%(c) 25%(d) 85%
5. The freezing point (0
0
C) of a solution containing 0.1 g of K
3
[Fe(CN) 6
] (Mol. wt. 329) in 100 g of water (K
f
= 1.86 K
kg mol
—1
) is(IIT 2011)(a) -2
-2.3 x10
(b) -2
-5.7 x10
(c) -3
-5.7 x10
(d) -2
-1.2 x10
6. A 0.2 molal aqueous solution of a weak acid(HX) is 20% ionised. The freezing point of the solution is (IIT 1995)(a) —0.45
0
C(b) 90
0
C(c) —0.31
0
C(d) —0.53
0
C
7. The molecular weight of benzoic acid in benzene as determined by depression in freezing point method
corresponds to    (IIT 1996)(a) ionization of benzoic acid(b) dimerisation of benzoic acid
(c) trimerization of benzoic acid(d) solvation of benzoic acid
8. The van’t Hoff factor of 0.1 M Ba(NO
3
) 2
solution is 2.74. The degree of dissociation kilogram of water, is(IIT 1999)(a) 91.3%(b) 87%(c) 100%(d) 74%
9. The vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985
N/m
2
. The vapour pressure of water is 3000 N/m
2
. The molecular mass of the solute is(IIT 1993)(a) 60(b) 120(c) 180(d) 380
10. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to (IIT 1983)(a) concentraton of solute in g/L(b) concentration of solute in g/100 mL
(c) mole fraction of solute(d) mole fraction of solvent
DPP : 05 / Solutions
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floor,Sector -15, Panchkula
9217610408,8699438881,8699438882
Page No.2
without ability to achieve it.
G od has never given dreams
11. Calculate the molal depression constant of a solvent which has freezing point 16.6
0
C and latent heat of
fusion 180.75 J g
—1
[JEE (Orissa) 2005]
(a) 2.68(b) 3.86(c) 4.68(d) 2.86
12. The elevation in boiling point for 13.44g of CuCl
2
dissolved in 1kg of water as solvent will be      (IIT 2005)(a) 0.05(b) 0.10(c) 0.16(d) 0.20
13. The amount of urea to be dissolved in 500 cc of water (K
f
= 1.86) to produce a depression of 0.186
0
C in the freezing
point is          [UGET (Manipal) 2006]
(a) 9 g(b) 6 g(c) 3 g(d) 0.3 g
14. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of
propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same
temperature will be(AIEEE 2007)(a) 360(b) 350(c) 300(d) 700
15. X is a non-volatile solute and Y is volatile solvent. The following vapour pressures are obtained by dissolving X and Y .
X mol L
—1
Y/mm Hg
0.1      P
1
0.25      P
2
0.01      P
3
The correct order of  vapour pressure is [EAMCET (Engg) 2010]
(a) P
1
< P
2
< P
3
(b) P
3
< P
2
< P
1
(c) P
3
< P
1
< P
2
(d) P
2
< P
1
< P
3
16. 18 g glucose glucose (C
6
H
12
O
6
) is added to 178.2g of water. The vapour pressure of water for this aqueous
solution at 100
0
C is(AIEEE 2006)(a) 759 torr(b) 7.60 torr(c) 76 torr(d) 752.4 torr
17. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol
—1
) in the same
solvent. If the denisities of both the solutions are assumed to be equal to 1 g cm
—3
, molar mass of the substance
will be(AIEEE 2007)(a) 210.0 g mol
—1
(b) 90.0 g mol
—1
(c) 115.0 g mol
—1
(d) 105.0g mol
—1
18. At 80
0
C, the vapour pressure of pure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg.
If a mixture of solution A and B boils at 80
0
C and 1 atm pressure, the amount of A in the mixture is(1atm = 760 mm
of Hg)(AIEEE 2008)(a) 50 mol %(b) 52 mol %(c) 34 mol %(d) 48 mol %
19. K
f
for water is 1.86 K kg mol
—1
. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol
(C
2
H
6
O
2
) must you add to get the freezing point of the solution lowered to —2.8
0
C ?(AIEEE 2012)(a) 27 g(b) 72 g(c) 93 g(d) 39 g
DPP : 05 / Solutions
SCF-07, 1
st
floor,Sector -15, Panchkula
9217610408,8699438881,8699438882
Page No.3
without ability to achieve it.
G od has never given dreams
20. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3
mol of Y is 550 mm Hg. At the same temperature if 1 mol of Y is further added to this solution, vapour pressure of
the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be respectively
(AIEEE 2009)(a) 200 and 300(b) 300 and 400(c) 400 and 600(d) 500 and 600
Integer based Questions
21. A compound H
2
X with molar mass of 80 g mol
—1
is dissolved in a solvent having density 0.4 g L
—1
. Assuming no
change in volume upon dissolution, the molality of a 3.2 molar solution is .... mol kg
—1
. (2014) 22. MX
2
dissociates into M
2+
and X
—
ions in an aqueous solution, with a degree of dissociation (a) is 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 ...... (2014) 23. If the freezing point of a 0.01 molal solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong
electrolyte) is –0.0558
0
C, the number of chloride(s) in the coordination sphere of the complex is ......(2015) [K
f
of water = 1.86 K kg mol
—1
]
1.(a) 2.(b) 3.(a) 4.(a) 5.(a) 6.(a) 7.(b) 8.(b) 9.(c) 10.(c) 11.(b) 12.(c) 13.(c) 14.(b) 15.(d) 16.(d) 17.(a) 18.(a) 19.(c) 20.(c) 21. 8 22. 2 23. 1
Page 4

DPP : 05 / Solutions
SCF-07, 1
st
floor,Sector -15, Panchkula
9217610408,8699438881,8699438882
Page No.1
without ability to achieve it.
G od has never given dreams
SOLUTIONS
DAILY  PRACTICE  PROBLEMS - 5
NON-MEDICAL ENTRANCE PREVIOUS YEAR QUESTIONS
1. The henry’s law constant for the solubility of N
2
gas in water at 298 K is 1.0 x 10
5
atm. The mole fraction of N
2
in air
is 0.8. The number of moles of N
2
from air dissolved in 10 moles of water at 298 K and 5 atm pressure is(IIT 2009)(a) -4
4.0 x10(b) -5
4.0 x10(c) -4
5.0 x10(d) -6
4.0 x10
2. The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its
aqueous solution at 27
0
C is 25 g mol
—1
. Therefore, its ionization percentage in this solution is(J & K 2009)(a) 75(b) 60(c) 80(d) 70
3. The freezing point of water is depressed by 0.37
0
C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal
solution of urea is depressed by(WBJEE 2008)(a) 0.37
0
C(b) 0.74
0
C(c) 0.185
0
C(d) 0
0
C
4. 0.004 M Na
2
SO
4
is isotonic with 0.01 M glucose. Degree of dissociation of Na
2
SO
4
is(IIT 2004)(a) 75%(b) 50%(c) 25%(d) 85%
5. The freezing point (0
0
C) of a solution containing 0.1 g of K
3
[Fe(CN) 6
] (Mol. wt. 329) in 100 g of water (K
f
= 1.86 K
kg mol
—1
) is(IIT 2011)(a) -2
-2.3 x10
(b) -2
-5.7 x10
(c) -3
-5.7 x10
(d) -2
-1.2 x10
6. A 0.2 molal aqueous solution of a weak acid(HX) is 20% ionised. The freezing point of the solution is (IIT 1995)(a) —0.45
0
C(b) 90
0
C(c) —0.31
0
C(d) —0.53
0
C
7. The molecular weight of benzoic acid in benzene as determined by depression in freezing point method
corresponds to    (IIT 1996)(a) ionization of benzoic acid(b) dimerisation of benzoic acid
(c) trimerization of benzoic acid(d) solvation of benzoic acid
8. The van’t Hoff factor of 0.1 M Ba(NO
3
) 2
solution is 2.74. The degree of dissociation kilogram of water, is(IIT 1999)(a) 91.3%(b) 87%(c) 100%(d) 74%
9. The vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985
N/m
2
. The vapour pressure of water is 3000 N/m
2
. The molecular mass of the solute is(IIT 1993)(a) 60(b) 120(c) 180(d) 380
10. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to (IIT 1983)(a) concentraton of solute in g/L(b) concentration of solute in g/100 mL
(c) mole fraction of solute(d) mole fraction of solvent
DPP : 05 / Solutions
SCF-07, 1
st
floor,Sector -15, Panchkula
9217610408,8699438881,8699438882
Page No.2
without ability to achieve it.
G od has never given dreams
11. Calculate the molal depression constant of a solvent which has freezing point 16.6
0
C and latent heat of
fusion 180.75 J g
—1
[JEE (Orissa) 2005]
(a) 2.68(b) 3.86(c) 4.68(d) 2.86
12. The elevation in boiling point for 13.44g of CuCl
2
dissolved in 1kg of water as solvent will be      (IIT 2005)(a) 0.05(b) 0.10(c) 0.16(d) 0.20
13. The amount of urea to be dissolved in 500 cc of water (K
f
= 1.86) to produce a depression of 0.186
0
C in the freezing
point is          [UGET (Manipal) 2006]
(a) 9 g(b) 6 g(c) 3 g(d) 0.3 g
14. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of
propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same
temperature will be(AIEEE 2007)(a) 360(b) 350(c) 300(d) 700
15. X is a non-volatile solute and Y is volatile solvent. The following vapour pressures are obtained by dissolving X and Y .
X mol L
—1
Y/mm Hg
0.1      P
1
0.25      P
2
0.01      P
3
The correct order of  vapour pressure is [EAMCET (Engg) 2010]
(a) P
1
< P
2
< P
3
(b) P
3
< P
2
< P
1
(c) P
3
< P
1
< P
2
(d) P
2
< P
1
< P
3
16. 18 g glucose glucose (C
6
H
12
O
6
) is added to 178.2g of water. The vapour pressure of water for this aqueous
solution at 100
0
C is(AIEEE 2006)(a) 759 torr(b) 7.60 torr(c) 76 torr(d) 752.4 torr
17. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol
—1
) in the same
solvent. If the denisities of both the solutions are assumed to be equal to 1 g cm
—3
, molar mass of the substance
will be(AIEEE 2007)(a) 210.0 g mol
—1
(b) 90.0 g mol
—1
(c) 115.0 g mol
—1
(d) 105.0g mol
—1
18. At 80
0
C, the vapour pressure of pure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg.
If a mixture of solution A and B boils at 80
0
C and 1 atm pressure, the amount of A in the mixture is(1atm = 760 mm
of Hg)(AIEEE 2008)(a) 50 mol %(b) 52 mol %(c) 34 mol %(d) 48 mol %
19. K
f
for water is 1.86 K kg mol
—1
. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol
(C
2
H
6
O
2
) must you add to get the freezing point of the solution lowered to —2.8
0
C ?(AIEEE 2012)(a) 27 g(b) 72 g(c) 93 g(d) 39 g
DPP : 05 / Solutions
SCF-07, 1
st
floor,Sector -15, Panchkula
9217610408,8699438881,8699438882
Page No.3
without ability to achieve it.
G od has never given dreams
20. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3
mol of Y is 550 mm Hg. At the same temperature if 1 mol of Y is further added to this solution, vapour pressure of
the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be respectively
(AIEEE 2009)(a) 200 and 300(b) 300 and 400(c) 400 and 600(d) 500 and 600
Integer based Questions
21. A compound H
2
X with molar mass of 80 g mol
—1
is dissolved in a solvent having density 0.4 g L
—1
. Assuming no
change in volume upon dissolution, the molality of a 3.2 molar solution is .... mol kg
—1
. (2014) 22. MX
2
dissociates into M
2+
and X
—
ions in an aqueous solution, with a degree of dissociation (a) is 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 ...... (2014) 23. If the freezing point of a 0.01 molal solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong
electrolyte) is –0.0558
0
C, the number of chloride(s) in the coordination sphere of the complex is ......(2015) [K
f
of water = 1.86 K kg mol
—1
]
1.(a) 2.(b) 3.(a) 4.(a) 5.(a) 6.(a) 7.(b) 8.(b) 9.(c) 10.(c) 11.(b) 12.(c) 13.(c) 14.(b) 15.(d) 16.(d) 17.(a) 18.(a) 19.(c) 20.(c) 21. 8 22. 2 23. 1
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MEDICAL ENTRANCE PREVIOUS YEAR QUESTIONS
1. Camphor is used as solvent to determine molecular weight of non-volatile solute by Rast method because
camphor [AIPMT 2004]
(a) is readily available(b) is volatile
(c) molal depression constant is high(d) is solvent for organic substances
2. From the colligative properties of solution which one is the best method for the detremination of molecular weight
of proteins and polymers [AIPMT 2000]
(a) osmotic pressure(b) lowering in vapour pressure
(c) lowering in freezing point(d) elevation in boiling point
3. The solubility of a gas in water depends on [MP (PMT) 2002]
(a) nature of the gas(b) temperature(c) pressure of the gas(d) all of these
4. The van’t Hoff factor “i” for a compound which undergoes dissociation in one solvent and association in other
solvent is respectively [AIPMT (P) 2011]
(a) greater then one and greater than one(b) less than one and greater than one
(c) less than one and less than one(d) greater than one and less than one
5. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20
0
C are 440
mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in vapour phase would be
[AIPMT 2005]
(a) 0.786(b) 0.549(c) 0.478(d)      0.200
6. A 0.5 molal aqueous solution of a weak acid (HX) is 20 per cent ionized. The lowering in freezing point of this
solution is (K
f
= 1.86 K/m for water) [AIPMT 2007]
(a) 0.56 K(b) —0.56 K(c) 1.12 K(d) —1.12 K
7. A solution of urea (mol. mass = 56) boils at 100.18
0
C at atmospheric pressure. If K
f
and K
b
for water are 1.86 and
0.512 K molality
—1
respectively, the above solution will freeze at [AIPMT 2005]
(a) —6.54
0
C(b) 6.54
0
C(c) —0.654
0
C(d) 0.654
0
C
8. A solution containing 10 g per dm
3
of urea (mol. mass = 60) is isotonic with a 5% (wt. by vol.) solution of a non-
volatile solute. The molecular mass (in g mol
—1
) of  non-volatile solute is [AIPMT 2006]
(a) 350(b) 200(c) 250(d) 300
9. 1.0 g of a non-electrolyte solute (molar mass 250 g mol
—1
) was dissolved in 51.2 g of benzene. If the freezing point
depression constant of benzene is 5.12 K kg mol
—1
, the lowering in freezing point will be [AIPMT 2006]
(a) 0.5 K(b) 0.2 K(c) 0.4 K(d) 0.3 K
Page 5

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SOLUTIONS
DAILY  PRACTICE  PROBLEMS - 5
NON-MEDICAL ENTRANCE PREVIOUS YEAR QUESTIONS
1. The henry’s law constant for the solubility of N
2
gas in water at 298 K is 1.0 x 10
5
atm. The mole fraction of N
2
in air
is 0.8. The number of moles of N
2
from air dissolved in 10 moles of water at 298 K and 5 atm pressure is(IIT 2009)(a) -4
4.0 x10(b) -5
4.0 x10(c) -4
5.0 x10(d) -6
4.0 x10
2. The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its
aqueous solution at 27
0
C is 25 g mol
—1
. Therefore, its ionization percentage in this solution is(J & K 2009)(a) 75(b) 60(c) 80(d) 70
3. The freezing point of water is depressed by 0.37
0
C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal
solution of urea is depressed by(WBJEE 2008)(a) 0.37
0
C(b) 0.74
0
C(c) 0.185
0
C(d) 0
0
C
4. 0.004 M Na
2
SO
4
is isotonic with 0.01 M glucose. Degree of dissociation of Na
2
SO
4
is(IIT 2004)(a) 75%(b) 50%(c) 25%(d) 85%
5. The freezing point (0
0
C) of a solution containing 0.1 g of K
3
[Fe(CN) 6
] (Mol. wt. 329) in 100 g of water (K
f
= 1.86 K
kg mol
—1
) is(IIT 2011)(a) -2
-2.3 x10
(b) -2
-5.7 x10
(c) -3
-5.7 x10
(d) -2
-1.2 x10
6. A 0.2 molal aqueous solution of a weak acid(HX) is 20% ionised. The freezing point of the solution is (IIT 1995)(a) —0.45
0
C(b) 90
0
C(c) —0.31
0
C(d) —0.53
0
C
7. The molecular weight of benzoic acid in benzene as determined by depression in freezing point method
corresponds to    (IIT 1996)(a) ionization of benzoic acid(b) dimerisation of benzoic acid
(c) trimerization of benzoic acid(d) solvation of benzoic acid
8. The van’t Hoff factor of 0.1 M Ba(NO
3
) 2
solution is 2.74. The degree of dissociation kilogram of water, is(IIT 1999)(a) 91.3%(b) 87%(c) 100%(d) 74%
9. The vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985
N/m
2
. The vapour pressure of water is 3000 N/m
2
. The molecular mass of the solute is(IIT 1993)(a) 60(b) 120(c) 180(d) 380
10. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to (IIT 1983)(a) concentraton of solute in g/L(b) concentration of solute in g/100 mL
(c) mole fraction of solute(d) mole fraction of solvent
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11. Calculate the molal depression constant of a solvent which has freezing point 16.6
0
C and latent heat of
fusion 180.75 J g
—1
[JEE (Orissa) 2005]
(a) 2.68(b) 3.86(c) 4.68(d) 2.86
12. The elevation in boiling point for 13.44g of CuCl
2
dissolved in 1kg of water as solvent will be      (IIT 2005)(a) 0.05(b) 0.10(c) 0.16(d) 0.20
13. The amount of urea to be dissolved in 500 cc of water (K
f
= 1.86) to produce a depression of 0.186
0
C in the freezing
point is          [UGET (Manipal) 2006]
(a) 9 g(b) 6 g(c) 3 g(d) 0.3 g
14. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of
propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same
temperature will be(AIEEE 2007)(a) 360(b) 350(c) 300(d) 700
15. X is a non-volatile solute and Y is volatile solvent. The following vapour pressures are obtained by dissolving X and Y .
X mol L
—1
Y/mm Hg
0.1      P
1
0.25      P
2
0.01      P
3
The correct order of  vapour pressure is [EAMCET (Engg) 2010]
(a) P
1
< P
2
< P
3
(b) P
3
< P
2
< P
1
(c) P
3
< P
1
< P
2
(d) P
2
< P
1
< P
3
16. 18 g glucose glucose (C
6
H
12
O
6
) is added to 178.2g of water. The vapour pressure of water for this aqueous
solution at 100
0
C is(AIEEE 2006)(a) 759 torr(b) 7.60 torr(c) 76 torr(d) 752.4 torr
17. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol
—1
) in the same
solvent. If the denisities of both the solutions are assumed to be equal to 1 g cm
—3
, molar mass of the substance
will be(AIEEE 2007)(a) 210.0 g mol
—1
(b) 90.0 g mol
—1
(c) 115.0 g mol
—1
(d) 105.0g mol
—1
18. At 80
0
C, the vapour pressure of pure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg.
If a mixture of solution A and B boils at 80
0
C and 1 atm pressure, the amount of A in the mixture is(1atm = 760 mm
of Hg)(AIEEE 2008)(a) 50 mol %(b) 52 mol %(c) 34 mol %(d) 48 mol %
19. K
f
for water is 1.86 K kg mol
—1
. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol
(C
2
H
6
O
2
) must you add to get the freezing point of the solution lowered to —2.8
0
C ?(AIEEE 2012)(a) 27 g(b) 72 g(c) 93 g(d) 39 g
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20. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3
mol of Y is 550 mm Hg. At the same temperature if 1 mol of Y is further added to this solution, vapour pressure of
the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be respectively
(AIEEE 2009)(a) 200 and 300(b) 300 and 400(c) 400 and 600(d) 500 and 600
Integer based Questions
21. A compound H
2
X with molar mass of 80 g mol
—1
is dissolved in a solvent having density 0.4 g L
—1
. Assuming no
change in volume upon dissolution, the molality of a 3.2 molar solution is .... mol kg
—1
. (2014) 22. MX
2
dissociates into M
2+
and X
—
ions in an aqueous solution, with a degree of dissociation (a) is 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 ...... (2014) 23. If the freezing point of a 0.01 molal solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong
electrolyte) is –0.0558
0
C, the number of chloride(s) in the coordination sphere of the complex is ......(2015) [K
f
of water = 1.86 K kg mol
—1
]
1.(a) 2.(b) 3.(a) 4.(a) 5.(a) 6.(a) 7.(b) 8.(b) 9.(c) 10.(c) 11.(b) 12.(c) 13.(c) 14.(b) 15.(d) 16.(d) 17.(a) 18.(a) 19.(c) 20.(c) 21. 8 22. 2 23. 1
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MEDICAL ENTRANCE PREVIOUS YEAR QUESTIONS
1. Camphor is used as solvent to determine molecular weight of non-volatile solute by Rast method because
camphor [AIPMT 2004]
(a) is readily available(b) is volatile
(c) molal depression constant is high(d) is solvent for organic substances
2. From the colligative properties of solution which one is the best method for the detremination of molecular weight
of proteins and polymers [AIPMT 2000]
(a) osmotic pressure(b) lowering in vapour pressure
(c) lowering in freezing point(d) elevation in boiling point
3. The solubility of a gas in water depends on [MP (PMT) 2002]
(a) nature of the gas(b) temperature(c) pressure of the gas(d) all of these
4. The van’t Hoff factor “i” for a compound which undergoes dissociation in one solvent and association in other
solvent is respectively [AIPMT (P) 2011]
(a) greater then one and greater than one(b) less than one and greater than one
(c) less than one and less than one(d) greater than one and less than one
5. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20
0
C are 440
mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in vapour phase would be
[AIPMT 2005]
(a) 0.786(b) 0.549(c) 0.478(d)      0.200
6. A 0.5 molal aqueous solution of a weak acid (HX) is 20 per cent ionized. The lowering in freezing point of this
solution is (K
f
= 1.86 K/m for water) [AIPMT 2007]
(a) 0.56 K(b) —0.56 K(c) 1.12 K(d) —1.12 K
7. A solution of urea (mol. mass = 56) boils at 100.18
0
C at atmospheric pressure. If K
f
and K
b
for water are 1.86 and
0.512 K molality
—1
respectively, the above solution will freeze at [AIPMT 2005]
(a) —6.54
0
C(b) 6.54
0
C(c) —0.654
0
C(d) 0.654
0
C
8. A solution containing 10 g per dm
3
of urea (mol. mass = 60) is isotonic with a 5% (wt. by vol.) solution of a non-
volatile solute. The molecular mass (in g mol
—1
) of  non-volatile solute is [AIPMT 2006]
(a) 350(b) 200(c) 250(d) 300
9. 1.0 g of a non-electrolyte solute (molar mass 250 g mol
—1
) was dissolved in 51.2 g of benzene. If the freezing point
depression constant of benzene is 5.12 K kg mol
—1
, the lowering in freezing point will be [AIPMT 2006]
(a) 0.5 K(b) 0.2 K(c) 0.4 K(d) 0.3 K
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10. A 0.0020 m aqueous solution of an ionic compound Co(NH
3
) 5
(NO
2
)Cl freezes at —0.00732
0
C. Number of moles
of ions which 1 mol of ionic compound produces on being dissolved in water will be (K
f
= +1.86
0
C/m) [AIPMT 2009]
(a) 1(b) 2(c) 3(d) 4
11. A solution of sucrose (molar mass 342 g mol
—1
) has been produced by dissolving 68.5 g sucrose in 1000 g water.
The freezing point of the solution obtained will be (K
f
for H
2
O = 1.86 K kg mol
—1
) [AIPMT (P) 2010]
(a) —0.372
0
C(b) —0.520
0
C(c) +0.372
0
C(d) —0.570
0
C
12. The freezing point depression constant for water is —1.86
0
Cm
—1
. If 5.00 g Na
2
SO
4
is dissolved in 45.0 g H
2
O, the
freezing point is changed by —3.82
0
C, Calculate the van’t Hoff factor for Na
2
SO
4
[AIPMT (P) 2011]
(a) 0.381(b) 2.05(c) 2.63(d) 3.11
13. p
A
and p
B
are the vapour pressure of pure liquid components. A and B, respectively of an ideal binary solution. If x
A
represents the mole fraction of component A, the total pressure of the solution will be [AIPMT (P) 2012]
(a) p
B
+ x
A
(p
B
—p
A
)(b) p
B
+ x
A
(p
A
—p
B
)(c) p
A
+ x
A
(p
B
—p
A
)(d) p
A
+ x
A
(p
A
—p
B
) 14. Which of the following compounds can be used as antifreeze in automobile radiators ? [AIPMT (M) 2012]
(a) Methyl alcohol(b) Glycol(c) Nitrophenol(d) Ethyl alcohol
15. 200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is
found to be 2.57 x 10
—3
bar. The molar mass of protein will be (R = 0.083 L bar mol
—1
) [AIPMT Mains 2011]
(a) 51022 g mol
—1
(b)         1211g mol
—1
(c) 31011 g mol
—1
(d)         61038g mol
—1
16. Vapour pressure of chloroform() 3
CHCl and dichloromethane() 2 2
CH Cl at 25
0
C are 200 mm of Hg and 415mm Hg
respectively, vapour pressure of the solution obtained by mixing 25.5 g
3
CHCl and 40 g of
2 2
CH Cl at the same
temperature will be [AIPMT (M) 2012]
(a) 347.9 mm Hg(b) 285 mm Hg(c) 173.9 mm Hg(d) 615.0 mm Hg
17. The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the
solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the
decrease in the vapour pressure is to be 20 mm of mercury ? [CBSE 1998]
(a) 0.8(b) 0.6(c) 0.4(d) 0.2
18. The vapour pressure of a solvent A is 0.80 atm. When a non-volatile substance B is added to this solvent, its vapour
pressure drops to 0.6 atm. The mole fraction of B in the solution is [PMT (MP) 2000]
(a) 0.25(b) 0.50(c) 0.75(d) 0.90
19. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolytic
solid, weighing 2.175 g, is added to 39.08 g of benzene. The vapour pressure of the solution is 600 mm of Hg.
What is the molecular mass of the solid substance ? [CBSE 1999]
(a) 49.50(b) 59.6(c) 69.5(d) 79.8
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