Solutions Practice Questions - DPP for JEE

 Page 1


1. (b) According to Le-chateliers principle, for an exothermic reaction
(??< 0) increase in temperature decreases the solubility.
2. (b) =  + 
= 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg
The observed   is 100 mm Hg which is less than 104 mm Hg.
Hence the solution shows negative deviation.
3. (d) According to Henry’s law,
m = k × p
given K
H
 = 1.4 × 10
–3
 = 0.5 or
No. of moles; 
4. (c) These two components A and B follows the condition of Raoult’s
law if the force of attraction between A and B is equal to the force
of attraction between A and A or B and B.
5. (c) There is no change in vapour pressure.
6. (b)
 = 
This is condition for ideal solution.
7. (d) C
2
H
5
I and C
2
H
5
OH form non-ideal solution.
Page 2


1. (b) According to Le-chateliers principle, for an exothermic reaction
(??< 0) increase in temperature decreases the solubility.
2. (b) =  + 
= 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg
The observed   is 100 mm Hg which is less than 104 mm Hg.
Hence the solution shows negative deviation.
3. (d) According to Henry’s law,
m = k × p
given K
H
 = 1.4 × 10
–3
 = 0.5 or
No. of moles; 
4. (c) These two components A and B follows the condition of Raoult’s
law if the force of attraction between A and B is equal to the force
of attraction between A and A or B and B.
5. (c) There is no change in vapour pressure.
6. (b)
 = 
This is condition for ideal solution.
7. (d) C
2
H
5
I and C
2
H
5
OH form non-ideal solution.
8. (b) For this solution intermolecular interactions between n-heptane
and ethanol are weaker than n-heptane-n-heptane & ethanol-
ethanol interactions hence the solution of n-heptane and ethanol is
non-ideal and shows positive deviation from Raoult’s law.
9. (a) Liquid solvent and solid solvent are in equilibrium.
10. (b) The molality involves weights of the solute and the solvent. Since
weight does not change with the temperature, therefore molality
does not depend upon the temperature.
11. (d) Molarity (M) = 
wt. =  g
wt. of 70% acid = = 45 g
12. (c) Vant Hoff factor i = 4 in case of (NH
4
)
3
PO
4
,
(NH
4
)
3
 PO
4
 
 
13. (d) ?T
b
 = iK
b
 m
Given, (?T
b
)
X
 > (?T
b
)
Y
? i
X
 K
b
 m > i
Y
K
b
 m
(K
b
 is same for same solvent)
i
X
 > i
Y
So, x is undergoing dissociation in water.
14. (a) Depression in F.P. No. of particles.
 provides five ions on ionisation
while KCl provides two ions
Page 3


1. (b) According to Le-chateliers principle, for an exothermic reaction
(??< 0) increase in temperature decreases the solubility.
2. (b) =  + 
= 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg
The observed   is 100 mm Hg which is less than 104 mm Hg.
Hence the solution shows negative deviation.
3. (d) According to Henry’s law,
m = k × p
given K
H
 = 1.4 × 10
–3
 = 0.5 or
No. of moles; 
4. (c) These two components A and B follows the condition of Raoult’s
law if the force of attraction between A and B is equal to the force
of attraction between A and A or B and B.
5. (c) There is no change in vapour pressure.
6. (b)
 = 
This is condition for ideal solution.
7. (d) C
2
H
5
I and C
2
H
5
OH form non-ideal solution.
8. (b) For this solution intermolecular interactions between n-heptane
and ethanol are weaker than n-heptane-n-heptane & ethanol-
ethanol interactions hence the solution of n-heptane and ethanol is
non-ideal and shows positive deviation from Raoult’s law.
9. (a) Liquid solvent and solid solvent are in equilibrium.
10. (b) The molality involves weights of the solute and the solvent. Since
weight does not change with the temperature, therefore molality
does not depend upon the temperature.
11. (d) Molarity (M) = 
wt. =  g
wt. of 70% acid = = 45 g
12. (c) Vant Hoff factor i = 4 in case of (NH
4
)
3
PO
4
,
(NH
4
)
3
 PO
4
 
 
13. (d) ?T
b
 = iK
b
 m
Given, (?T
b
)
X
 > (?T
b
)
Y
? i
X
 K
b
 m > i
Y
K
b
 m
(K
b
 is same for same solvent)
i
X
 > i
Y
So, x is undergoing dissociation in water.
14. (a) Depression in F.P. No. of particles.
 provides five ions on ionisation
while KCl provides two ions
C
6
H
12
O
6 
and C
12
H
22
O
11 
are not ionised so they have single particle.
Hence, Al
2
(SO
4
)
3
 have maximum value of depression in F.P or lowest
F.P
15. (b) Sodium sulphate dissociates as
hence van’t hoff factor i = 3
Now 
= 3 × 1.86 × 0.01 = 0.0558 K
16. (a) and  both dissociates to give 5 ions or i
= 5
K
4
[Fe(CN)
6
]  4K
+
 + [Fe(CN)
6
]
4–
and 
17. (b) 1 molal solution means 1 mole of solute dissolved in 1000 gm
solvent.
? n
solute
 = 1        w
solvent
 = 1000g
? n
solvent
 =  = 55.56
X
solute
 =  = 0.0177
18. (b) Concentration of
 = 0.955 M
 = 2 × 0.955 = 1.91 M
 = 0.955 M
19. (d) Azeotrope of HCl + H
2
O contains 20.2% HCl.
20. (a) Mole fraction of any component A
Page 4


1. (b) According to Le-chateliers principle, for an exothermic reaction
(??< 0) increase in temperature decreases the solubility.
2. (b) =  + 
= 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg
The observed   is 100 mm Hg which is less than 104 mm Hg.
Hence the solution shows negative deviation.
3. (d) According to Henry’s law,
m = k × p
given K
H
 = 1.4 × 10
–3
 = 0.5 or
No. of moles; 
4. (c) These two components A and B follows the condition of Raoult’s
law if the force of attraction between A and B is equal to the force
of attraction between A and A or B and B.
5. (c) There is no change in vapour pressure.
6. (b)
 = 
This is condition for ideal solution.
7. (d) C
2
H
5
I and C
2
H
5
OH form non-ideal solution.
8. (b) For this solution intermolecular interactions between n-heptane
and ethanol are weaker than n-heptane-n-heptane & ethanol-
ethanol interactions hence the solution of n-heptane and ethanol is
non-ideal and shows positive deviation from Raoult’s law.
9. (a) Liquid solvent and solid solvent are in equilibrium.
10. (b) The molality involves weights of the solute and the solvent. Since
weight does not change with the temperature, therefore molality
does not depend upon the temperature.
11. (d) Molarity (M) = 
wt. =  g
wt. of 70% acid = = 45 g
12. (c) Vant Hoff factor i = 4 in case of (NH
4
)
3
PO
4
,
(NH
4
)
3
 PO
4
 
 
13. (d) ?T
b
 = iK
b
 m
Given, (?T
b
)
X
 > (?T
b
)
Y
? i
X
 K
b
 m > i
Y
K
b
 m
(K
b
 is same for same solvent)
i
X
 > i
Y
So, x is undergoing dissociation in water.
14. (a) Depression in F.P. No. of particles.
 provides five ions on ionisation
while KCl provides two ions
C
6
H
12
O
6 
and C
12
H
22
O
11 
are not ionised so they have single particle.
Hence, Al
2
(SO
4
)
3
 have maximum value of depression in F.P or lowest
F.P
15. (b) Sodium sulphate dissociates as
hence van’t hoff factor i = 3
Now 
= 3 × 1.86 × 0.01 = 0.0558 K
16. (a) and  both dissociates to give 5 ions or i
= 5
K
4
[Fe(CN)
6
]  4K
+
 + [Fe(CN)
6
]
4–
and 
17. (b) 1 molal solution means 1 mole of solute dissolved in 1000 gm
solvent.
? n
solute
 = 1        w
solvent
 = 1000g
? n
solvent
 =  = 55.56
X
solute
 =  = 0.0177
18. (b) Concentration of
 = 0.955 M
 = 2 × 0.955 = 1.91 M
 = 0.955 M
19. (d) Azeotrope of HCl + H
2
O contains 20.2% HCl.
20. (a) Mole fraction of any component A
As total no. of moles > no. of moles of A
thus x can never be equal to one or zero.
21. (50) = 0.3 °C
0.3 °C = 
0.3 = 
 = 50 g
The amount used should be more than 50 g.
22. (0.83) Let a  be the give of dissociation then
  
Van’t Hoff's factor i = 
Again Van’t Hoff’s factor
Equating to both values of i,   
23. (2) ?T
f
 = 0 – (– 0.00732°) = 0.00732
?T
f
 = i × K
f
 × m
24. (–0.654) As ?T
f
 = K
f
. m
?T
b
 = K
b
. m
Hence, we have  
Page 5


1. (b) According to Le-chateliers principle, for an exothermic reaction
(??< 0) increase in temperature decreases the solubility.
2. (b) =  + 
= 80.0 × 0.4 + 120.0 × 0.6 = 104 mm Hg
The observed   is 100 mm Hg which is less than 104 mm Hg.
Hence the solution shows negative deviation.
3. (d) According to Henry’s law,
m = k × p
given K
H
 = 1.4 × 10
–3
 = 0.5 or
No. of moles; 
4. (c) These two components A and B follows the condition of Raoult’s
law if the force of attraction between A and B is equal to the force
of attraction between A and A or B and B.
5. (c) There is no change in vapour pressure.
6. (b)
 = 
This is condition for ideal solution.
7. (d) C
2
H
5
I and C
2
H
5
OH form non-ideal solution.
8. (b) For this solution intermolecular interactions between n-heptane
and ethanol are weaker than n-heptane-n-heptane & ethanol-
ethanol interactions hence the solution of n-heptane and ethanol is
non-ideal and shows positive deviation from Raoult’s law.
9. (a) Liquid solvent and solid solvent are in equilibrium.
10. (b) The molality involves weights of the solute and the solvent. Since
weight does not change with the temperature, therefore molality
does not depend upon the temperature.
11. (d) Molarity (M) = 
wt. =  g
wt. of 70% acid = = 45 g
12. (c) Vant Hoff factor i = 4 in case of (NH
4
)
3
PO
4
,
(NH
4
)
3
 PO
4
 
 
13. (d) ?T
b
 = iK
b
 m
Given, (?T
b
)
X
 > (?T
b
)
Y
? i
X
 K
b
 m > i
Y
K
b
 m
(K
b
 is same for same solvent)
i
X
 > i
Y
So, x is undergoing dissociation in water.
14. (a) Depression in F.P. No. of particles.
 provides five ions on ionisation
while KCl provides two ions
C
6
H
12
O
6 
and C
12
H
22
O
11 
are not ionised so they have single particle.
Hence, Al
2
(SO
4
)
3
 have maximum value of depression in F.P or lowest
F.P
15. (b) Sodium sulphate dissociates as
hence van’t hoff factor i = 3
Now 
= 3 × 1.86 × 0.01 = 0.0558 K
16. (a) and  both dissociates to give 5 ions or i
= 5
K
4
[Fe(CN)
6
]  4K
+
 + [Fe(CN)
6
]
4–
and 
17. (b) 1 molal solution means 1 mole of solute dissolved in 1000 gm
solvent.
? n
solute
 = 1        w
solvent
 = 1000g
? n
solvent
 =  = 55.56
X
solute
 =  = 0.0177
18. (b) Concentration of
 = 0.955 M
 = 2 × 0.955 = 1.91 M
 = 0.955 M
19. (d) Azeotrope of HCl + H
2
O contains 20.2% HCl.
20. (a) Mole fraction of any component A
As total no. of moles > no. of moles of A
thus x can never be equal to one or zero.
21. (50) = 0.3 °C
0.3 °C = 
0.3 = 
 = 50 g
The amount used should be more than 50 g.
22. (0.83) Let a  be the give of dissociation then
  
Van’t Hoff's factor i = 
Again Van’t Hoff’s factor
Equating to both values of i,   
23. (2) ?T
f
 = 0 – (– 0.00732°) = 0.00732
?T
f
 = i × K
f
 × m
24. (–0.654) As ?T
f
 = K
f
. m
?T
b
 = K
b
. m
Hence, we have  
or  
= 0.18 ×  = 0.654°C
As the Freezing Point of pure water is 0 °C,
?T
f
 = 0 –T
f
0.654 = 0 – T
f
? T
f
 = – 0.654
Thus the freezing point of solution will be – 0.654 °C.
25. (0.052) ; 
 ?