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Solutions Part-02
Solutions DPP-01
1.
How many grams of HNO3 is required to prepare 400 mL solution of 0.2 M HNO3?
(1) 5.04 g
(2) 5040 g
(3) 25.2 g
(4) 2.52 g
2.
20.6 g NaBr is dissolved in 500 mL solution what is the molarity of resulting solution?
(1) 0.6
(2) 0.4
(3) 1
(4) 0.2
3.
Calculate molality of the solution obtained by dissolving 11.7 g NaCl in 500 g water
(1) 0.1 m
(2) 0.3 m
(3) 0.2 m
(4) 0.4 m
4.
Density of 2.03 M aqueous solution of acetic acid is 1.017 g mL –1 (molecular mass of acetic acid is 60).
Calculate the molality of solution?
(1) 2.27
(2) 1.27
(3) 3.27
(4) 4.27
5.
A molar solution is one that contains one mole of solute in
(1) 1000 g of the solvent
(2) one litre of the solution
(3) 1000 g of the solution
(4) 22.4 litres of the solution
6.
What is the normality of 2M H3PO2 solution?
(1) 0.5 N
(2) 1.0N
(3) 2.0 N
(4) 3.0 N
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Solutions Part-02
7.
What is the molarity of 1N H2SO4 solution?
(1) 1 M
(2) 2M
(3) 0.5 M
(4) 3M
8.
The normality of 3M H3PO4 is –
(1) 3
(2) 6
(3) 9
(4) 4
9.
171 g of cane sugar (C12H22O11) is dissolved in 1 litre of water. The molarity of the solution is
(1) 2.0 M
(2) 1.0 M
(3) 0.5 M
(4) 0.25 M
10.
If 5.6 g of KOH is present in 40 mL of solution. What is the normality of the solution [Molecular mass of
KOH = 56]
(1) 0.5
(2) 2.5
(3) 1
(4) 2
11.
With 63 gm of oxalic acid how many litres of
𝐍
𝟏𝟎
solution can be prepared
(1) 100 litre
(2) 10 litre
(3) 14 litre
(4) 1000 litre
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Solutions Part-02
Question
Answer
1
1
2
2
3
4
Answer Key
4
5
6
1
2
3
7
3
8
3
9
3
10
2
11
3
SOLUTIONS DPP-01
1.
n
Molarity =
v
w
Molarity =
63×0.4
w = 63 × 0.4 × 0.2
w = 5.04 g
2.
n
20.6
=
v 103  0.5
Molarity =
Molarity = 0.4 M
3.
Molality =
moles of solute
weight of solvent(Kg)
11.7
× 1000
58.5 × 500
= 0.4 M
=
1000M
4.
m=
5.
A molar solution means one mole of solute is present in one litre of solution.
6.
N = M × v.f.
1000d−MMw
1000 × 2.03
m=
(1000 × 1.017) − (2.03 × 60)
2030
2030
m=
=
= 2.27m
1017 − 121.8 895.2
N = 2 × 1 = 2N
7.
N = M × v.f.
1
M = = 0.5M
2
8.
Basicity of H3PO4 is 3
We know that N = M × n  Normality = 3 × 3 = 9 M
9.
Molarity =
w
171
=
= 0.5 M
molecular weight  volume in litre 342  1
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Solutions Part-02
10.
We know that
Molarity =
M=
Mass of solute × 1000
Molar Mass of solute × Volume of solution (mL)
5.6
 1000 = 2.5 M
56  40
Normality = 2.5 × 1
= 2.5 N
11.
Oxalic acid C2H2O4
w
1
N =( )× ×2
Mw
V
1
63 1
=
× ×2
10 90 V
63×10
V=
× 2 = 7 × 2 = 14
90
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Solutions Part-03
Solutions DPP-02
1.
23 g ethanol is dissolved in 36 g water. Find mole fraction of ethanol?
(1) 2
(2) 0.5
(3) 0.2
(4) 0.8
2.
Calculate normality of 2.1% (w/V) H2SO4 solution?
(1) 2.14 N
(2) 4.28 N
(3) 0.428 N
(4) 0.214 N
3.
Calculate the mole percentage of CH3OH and H2O respectively in 60% (by mass) aqueous solution of
CH3OH
(1) 45.8, 54.2
(2) 54.2, 45.8
(3) 50, 50
(4) 60, 40
4.
Find out the molarity of 60% (w/W) H2SO4 (density = 1.96 g/ml).
(1) 10 M
(2) 11 M
(3) 12 M
(4) 9 M
5.
Find the percentage by mass and mass fraction of insulin in the solution prepared by dissolving 4.48 g
of insulin in 26.52 g of water.
(1) 0.125
(2) 0.145
(3) 0.368
(4) 0.482
6.
A solution was prepared by adding 120 cm3 of C2H5OH to water until the volume of the solution was
190 cm3. Find volume percent of C2H5OH in the solution.
(1) 50.4
(2) 82.5
(3) 63.1
(4) 75.8
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Solutions Part-03
7.
Conc. H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. Its normality is
(1) 2 N
(2) 19.8 N
(3) 39.6 N
(4) 98 N
8
The mole fraction of the solute in one molal aqueous solution is
(1) 0.027
(2) 0.036
(3) 0.018
(4) 0.009
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Solutions Part-03
Question
Answer
1
3
2
3
Answer Key
3
4
1
3
5
2
6
3
7
3
8
3
SOLUTIONS DPP-02
1.
2.
23 1
= = 0.5
46 2
36
nH2O =
=2
18
nC2H5OH
0.5
0.5
X C2H5OH =
=
=
nC2H5OH + nH2 O 0.5 + 2 2.5
X C2H5OH = 0.2
nC2H5OH =
2.1 (w / v) H2SO4 means 2.1 g of H2SO4 is present in 100 mL of solution
Moles of H2SO4 =
2.1
98
gram eq
2.1 × 2
Normality =
=
= 0.428N
v(L)
98 × 0.1
3.
60
= 1.875
32
40
=
= 2.22
18
nCH3OH =
nH2O
Moles percent of CH3 OH =
1.875
× 100%
1.875 + 2.22
= 45.8%
Mole percent of H2O = 100 – 45.8 = 54.2%
4.
Molarity =
=
5.
Moles of solute
Mass in gram  Density  1000
=
Volume of solution (L) Molar mass  mass of solution (g)
60 × 1.96 × 1000
= 12 M
98 × 100
Mass of solution = 4.48 + 26.52 = 31g
Mass fraction=
6.
4.48
31
= 0.145
Volume of solute = 120 cm3
Volume of solution = 190 cm3
120

Volume fraction =
= 0.631
190
120
Volume percent =
× 100 = 63.1%
190
7.
Strength of H2SO4 = 98 × 19.8 g/litre
S
98 × 19.8
S = eq. wt. × N; N =
=
= 39.6
eq.wt.
49
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Solutions Part-03
8.
W = 1000 gm (H2O); n = 1 mole
W 1000
N=
=
= 55.55
M
18
xSolute =
n
1
=
= 0.018.
n + N 1 + 55.55
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Solutions Part-04
Solutions DPP-03
1.
The molarity of a solution made by mixing 50 ml of conc. H2SO4 (36N) with 50 ml of water is
(1) 36 M
(2) 18 M
(3) 9 M
(4) 6 M
2.
Find the volume of water needed to be mixed with 10 ml 10N HNO3 to get 0.1 N HNO3 solution
(1) 1000 ml
(2) 990 ml
(3) 1010 ml
(4) 10 ml
3.
A 5 molar solution of H2SO4 is diluted from 1 litre to 10 litres. What is the normality of the solution
(1) 0.25 N
(2) 1 N
(3) 2 N
(4) 7 N
4.
25 ml of 3.0 M HNO3 are mixed with 75 ml of 4.0 M HNO3. If the volumes are additive, the molarity of
the final mixture would be
(1) 3.25 M
(2) 4.0 M
(3) 3.75 M
(4) 3.50 M
5.
If 1 M and 2.5 litre NaOH solution is mixed with another 0.5 M and 3 litre NaOH solution, then molarity
of the resultant solution will be
(1) 1.0 M
(2) 0.73 M
(3) 0.80 M
(4) 0.50 M
6.
The volumes of 4N HCl and 10 N HCl required to make 1 litre of 6N HCl are
(1) 0.75 litre of 10 N HCl and 0.25 litre of 4 N HCl
(2) 0.25 litre of 4 N HCl and 0.75 litre of 10 N HCl
(3) 0.67 litre of 4 N HCl and 0.33 litre of 10 N HCl
(4) 0.80 litre of 4 N HCl and 0.20 litre of 10 N HCl
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Solutions Part-04
7.
Hydrochloric acid solution A and B have concentration of 0.5N and 0.1N respectively. The volumes of
solutions A and B required to make 2 litres of 0.2 N HCl are
(1) 0.5 L of A + 1.5 L of B
(2) 1.5 L of A + 0.5 L of B
(3) 1.0 L of A + 1.0 L of B
(4) 0.75 L of A + 1.25 L of B
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Solutions Part-04
Question
Answer
Answer Key
2
3
4
2
2
3
1
3
5
2
6
3
7
1
SOLUTIONS DPP-03
1.
N1V1= N2V2, 36 × 50 = N2 × 100
36×50
N2 =
= 18; 18N H2SO4 = 9M H2SO4
100
2.
N1V1 = N2V2
10 × 10 = 0.1 (10 + V)
10×10
V=
– 10 = 1000 – 10 = 990 ml.
0.1
3.
N1V1 = N2V2
(5 × 2) × 1 = N2 × 10
N2 = 1 N
M1 V1 +M2 V2
(3×25)+(4×75)
4.
Mmix =
5.
(2.5 × 1 + 3 × 0.5) = M3 × 5.5
=
V1 +V2
25+75
75 + 300 375
Mmix
=
= 3.75M
100
100
or 2.5 + 1.5 = M3 ×5.5 or M3 =
6.
4
5.5
=0.73 M
N1V1 + N2V2 = NV
4x + 10 (1 – x) = 6 × 1; –6x =–4; x = 0.66
7.
Nmix =
N1 V1 +N2 V2
V1 +V2
0.5x + 0.1(2 − x)
0.2 =
2
0.4 = 0.5x + 0.2 – 0.1x
0.4x = 0.2
x = 0.5
0.5L of A
(2 – 0.5) L of B
1.5L of B
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Solutions Part-05
Solutions DPP-04
1.
The volume strength of 1.5 N H2O2 solution is
(1) 4.8
(2) 5.2
(3) 8.8
(4) 8.4
2.
Calculate volume strength of H2O2 in 8L solution which yields 120 L of O2 at NTP
(1) 5 V
(2) 30 V
(3) 15 V
(4) 10 V
3.
The normality of 10 mL of a '20 V' H2O2 solution is
(1) 1.79
(2) 3.58
(3) 60.86
(4) 6.086
4.
H2O2 solution used for hair bleaching is sold as a solution of approximately 5.0 g H 2O2 per 100 mL of the
solution. The molecular mass of H2O2 is 34. The molarity of this solution is approximately: (1) 0.15 M
(2) 1.5 M
(3) 3.0 M
(4) 3.4 M
5.
The concentration of a solution of H2O2 is 6.8%(w/V) then the volume concentration of the solution is:(1) 22.4
(2) 11.2
(3) 20
(4) 5
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Solutions Part-05
Question
Answer
Answer Key
1
2
3
4
3
2
4
2
5
1
SOLUTIONS DPP-04
1.
Volume strength = 1.5 × 5.6 = 8.4
2.
8L, H2O2 gives = 120 L O2
1L H2O2 gives = 15 L O2
i.e. 15 V H2O2 solution
3.
Volume strength = N × 5.6
20
N=
= 3.58
5.6
4.
Molarity =
5.
6.89 H2O2 in 100 mL solution
grameq
6.8 × 2
N=
=
=4
V(L)
34 × 0.1
nH2O2
V(L)
=
5
= 1.5 M
34  0.1
Volume strength = N × 5.6
= 4 × 5.6
= 22.4
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Solutions Part-07
Solutions DPP-05
1.
A and B form an ideal solution. Solution is prepared by mixing 40 g of A and 20 g of B at 300K. At 300K
𝐏𝐀𝟎 = 70 mm of Hg and 𝐏𝐁𝟎 = 50 mm of Hg.
Calculate vapour pressure of solution. (Molar mass of A = 20 g mol –1 & Molar mass of B = 40 g mol–1)
(1) 55
(2) 66
(3) 44
(4) 72
2.
1 mole heptane (V.P. = 92 mm of Hg) is mixed with 4 mole. Octane (V.P. = 31 mm of Hg), form an ideal
solution. Find out the vapour pressure of solution.
(1) 57.8
(2) 49.2
(3) 43.2
(4) 59.2
3.
At 88oC benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What
is the mole fraction of benzene in the mixture with toluene that will be boil at 88°C at 1 atm pressure
(benzene – toluene form an ideal solution).
(1) 0.52
(2) 0.64
(3) 0.62
(4) 0.74
4.
If 𝐏𝐀𝟎 and 𝐏𝐁𝟎 are 108 and 36 torr respectively. What will be the mole fraction of A in vapour phase if B has
mole fraction of 0.5 in solution:
(1) 0.25
(2) 0.75
(3) 0.60
(4) 0.35
5.
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at
20oC are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour
phase would be
(1) 0.549
(2) 0.200
(3) 0.786
(4) 0.478
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Solutions Part-07
Question
Answer
Answer Key
1
2
3
2
3
4
4
2
5
4
SOLUTIONS DPP-05
1.
nA =
40
=2
;
20
2 4
XA =
=
2.5 5
4
5
1
20
= 0.5
40
0.5 1
XB =
=
2.5 5
;
PA = XA . P0A =
PB = XB . P0B =
nB =
× 70 = 56 mm of Hg
× 50 = 10 mm of Hg
5
PS= PA + PB = 56 + 10
PS= 66 mm of Hg
2.
Total mole = 1 + 4 = 5
Mole fraction of heptane XA = 1/5
Mole fraction of octane XB = 4/5
PS = XAPA0 + XBPB0 =
3.
1
4
× 92 +
× 31= 43.2 mm of Hg.
5
5
PS = 760 torr, because solution boils at 88oC
PS = PB0 XB + Pt0 Xt
 760 = 900 a + 360 – 360 a
a = 0.74 where 'a' is mole fraction of C6H6(XB)
4.
XB = 0.5, XA = 1 – 0.5
PT = PA0XA + PB0 XB
= 108 × (0.5) + 36 (0.5)
= 54 + 18
= 72
YA =
5.
PA 0 XA 108 × 0.5
=
= 0.75
PT
72
n pentane
n hexane
=
Xpentane =
Ypentane =
1
4
1 4
1
1
= , Xhexane = 1 − =
1+4 5
5 5
440 
1
5
1
4
440  + 120 
5
5
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=
88
88
=
= 0.478
88 + 96 184
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Solutions Part-08
Solutions DPP-06
1.
The vapour pressure of benzene at 90o C is 1020 torr. A solution of 5 g of a solute in 58.5 g benzene has
vapour pressure 990 torr. The molecular weight of the solute is?
(1) 280
(2) 220
(3) 270
(4) 320
2.
The lowering of vapour pressure of a solvent by addition of a non-volatile solute to it, is directly
proportional to:
(1) Mole fraction of solute
(2) The nature of the solute in the solution
(3) The atmospheric pressure
(4) All
3.
The relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute. This
statement was given by:
(1) Raoult
(2) Henry
(3) Joule
(4) Dalton
4.
The vapour pressure of a solution having solid as solute and liquid as solvent is:
(1) Directly proportional to mole fraction of the solvent
(2) Inversely proportional to mole fraction of the solvent
(3) Directly proportional to mole fraction of the solute
(4) Inversely proportional to mole fraction of the solute
5.
One mole of non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution
relative to that of water is
2
(1)
3
1
(2)
3
1
(3)
2
3
(4)
2
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Solutions Part-08
Question
Answer
Answer Key
1
2
3
2
1
1
4
1
5
1
SOLUTIONS DPP-06
1.
2.
3.
P0 − PS w  M
1020 − 990
5  78
=

=
 m = 220
PS
m W
990
m  58.5
P0 −Ps
= XB
P0
P0 – Ps = P0XB
P0 −Ps
P0
= XB (Raoult’s law)
4.
PA = PA0XA
5.
Ps = P0Xsolvent
Ps
2
2
=
=
0
P
1+2 3
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Solutions Part-09
Solutions DPP-07
1.
0.15 g of a substance dissolved in 15 g of solvent boiled at a temperature higher by 0.216oC than that of
the pure solvent. What is the molar mass of the substance. [Kb for solvent = 2.16oCm–1]
(1) 125
(2) 80
(3) 100
(4) 120
2.
The rise in boiling point of a solution containing 1.8 g glucose in 100 g of a solvent is 0.1oC. The molal
elevation constant of the liquid is –
(1) 1.8
(2) 1
(3) 1.6
(4) 2.1
3.
The molal elevation constant is the ratio of the elevation in B.P. to:
(1) Molarity
(2) Molality
(3) Mole fraction of solute
(4) Mole fraction of solvent
4.
If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by 0.216oC than that of
the pure solvent. The molecular weight of the substance (molal elevation constant for the solvent is
2.16oC) is
(1) 1.01
(2) 10
(3) 10.1
(4) 100
5.
Elevation in boiling point was 0.52oC when 6 gm of a compound X was dissolved in 100 gm of water.
Molecular weight of X is (Kb for water is 0.52 per 1000 gm of water)
(1) 120
(2) 60
(3) 180
(4) 600
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Solutions Part-09
Question
Answer
Answer Key
1
2
3
3
2
2
4
4
5
2
SOLUTIONS DPP-07
1.
Given
Kb = 2.16oC, w = 0.15 g, Tb = 0.216oC, W = 15 g
w
0.15
ΔTb =
× 1000 × K b ⇒ 0.216 =
× 1000 × 2.16
m×W
m × 15
0.15 × 1000 × 2.16
⇒m=
= 100
0.216 × 15
2.
 Tb = Molality × Kb
w
ΔTb × m × W
ΔTb =
× 1000 × K b ⇒ K b =
m×W
1000 × w
o
given  Tb = 0.1 C, m = 180, W = 100, w = 1.8
180 × 0.1 × 100
Kb =
= 1.0° Cm–1
1000 × 1.8
3.
 Tb= Kb × m
⇒ Kb =
4.
ΔTb
m
Tb = Kb × m
0.15 1000
×
M
15
2.16 × 0.15 × 1000
M=
15 × 0.216
M = 100
0.126 = 2.16 ×
5.
Tb = Kb × m
0.52 = 0.52 ×
M=
6 1000
×
M 100
6 × 1000
= 60
100
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Solutions Part-10
Solutions DPP-08
1.
Molal depression constant for water is 1.86oC. The freezing point of a 0.05 molal solution of a nonelectrolyte in water is
(1) –1.86oC
(2) –0.93oC
(3) –0.093oC
(4) 0.93oC
2.
The amount of urea to be dissolved in 500 ml of water (K =18.6 K mole–1 in 100g solvent) to produce a
depression of 0.186oC in freezing point is
(1) 9 g
(2) 6 g
(3) 3 g
(4) 0.3 g
3.
The freezing point of a solution prepared from 1.25 gm of a non-electrolyte and 20 gm of water is
271.9 K. If molar depression constant is 1.86 K mole–1, then molar mass of the solute will be
(1) 105.7
(2) 106.7
(3) 115.3
(4) 93.9
4.
The molar freezing point constant for water is 1.86oC mole–1. If 342 gm of cane sugar (C12H22O11) are
dissolved in 1000 gm of water, the solution will freeze at
(1) –1.86oC
(2) 1.86oC
(3) –3.92oC
(4) 2.42oC
5.
After adding a solute freezing point of solution decreases to – 0.186 oC. Calculate Tb if Kf = 1.86 and
Kb = 0.521.
(1) 0.521 ºC
(2) 0.0521 ºC
(3) 1.86 ºC
(4) 0.0186 ºC
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Solutions Part-10
Answer Key
1
2
3
3
3
1
Question
Answer
4
1
5
2
SOLUTIONS DPP-08
1.
Tf = Kf × molality = 1.86 × 0.05 = 0.093oC
Thus freezing point = 0 – 0.093 = –0.093oC
2.
ΔTf =
100×K×w
m×W
0.186 =
100×18.6×w
60×500
w = 3g
3.
Molar mass =
Kf ×1000×w
ΔTf ×W
=
1.86×1000×1.25
20×1.1
= 105.68 = 105.7
4.
342
Tf = 1.86 ×(
)= 1.86o;  Tf = –1.86oC
342
5.
Tf = Kf × m  0.186 × m
So, m = 0.1, Put the value of m in Tb = Kb × m
Tb = 0.521 × (0.1) = 0.0521
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Solutions Part-11
Solutions DPP-09
1.
If 3 gm of glucose (mol. wt. = 180) is dissolved in 60 mL of water at 15oC. Then the osmotic pressure of
this solution will be
(1) 0.34 atm
(2) 0.65 atm
(3) 6.57 atm
(4) 5.57 atm
2.
The concentration in gms per litre of a solution of cane sugar (M = 342) which is isotonic with a solution
containing 6 gms of urea (M = 60) per litre is
(1) 3.42
(2) 34.2
(3) 5.7
(4) 19
3.
A solution contains non-volatile solute of molecular mass MP. Which of the following can be used to
calculate molecular mass of the solute in terms of osmotic pressure (m = Mass of solute, V = Volume of
solution and  = Osmotic pressure)
m
(1) Mp =   VRT

 m  RT
(2) Mp =  
V 
m 
(3) Mp =  
 V  RT
m
(4) Mp =   πRT
V
4.
Two solutions A and B are separated by semi- permeable membrane. If liquid flows from A to B then
(1) A is less concentrated than B
(2) A is more concentrated than B
(3) Both have same concentration
(4) None of these
5.
A 5% solution of cane sugar (mol. wt.= 342) is isotonic with 1% solution of a substance X. The molecular
weight of X is
(1) 34.2
(2) 171.2
(3) 68.4
(4) 136.8
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Solutions Part-11
Answer Key
1
2
3
3
2
2
Question
Answer
4
1
5
3
SOLUTIONS DPP-09
3×1000
1.
 = CRT =
2.
For isotonic solution
3.
π=
4.
Osmosis occur from dilute solution to concentrated solution. Therefore, solution A is less concentrated than B.
5.
Molar concentration of cane sugar =
180×60
× 0.0821 × 288 = 6.56 atm
w1
=
w2
m1 V1 m2 V2
w1
6
342 × 6
=
=
= 34.2
342 × 1 60 × 1
60
n
 m  RT
RT ⇒ MP =  
V
V 
Molar concentration of X =
5 1000 50

=
342 100 342
1 1000 10

=
m 100 m
10 50
or m = 68.4
=
m 342
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Solutions Part-12
Solutions DPP-10
1.
Calculate the percentage degree of dissociation of an electrolyte AB 2 (normal molar mass = 164) in H 2O
if observed molar mass is 65.6
(1) 50%
(2) 25%
(3) 75%
(4) None
2.
Van't Hoff factor of Hg2Cl2 in its aqueous solution will be (Hg2Cl2 is 80% ionized in the solution)
(1) 1.6
(2) 2.6
(3) 3.6
(4) 4.6
3.
The Van’t Hoff factor (i) for a dilute solution of K3[Fe(CN)6] is :
(1) 10
(2) 4
(3) 5
(4) 0.25
4.
The degree of dissociation () of a weak electrolyte, AxBy is related to Vant Hoff factor (i) by the
expression:
5.
(1) α =
x + y −1
i −1
(2) α =
x + y +1
i −1
(3) α =
i −1
(x + y − 1)
(4) α =
i −1
x + y +1
The Vant Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74 . The degree of dissociation is:
(1) 91.3%
(2) 87%
(3) 100%
(4) 74%
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TG: @Chalnaayaaar
Solutions Part-12
Question
Answer
Answer Key
1
2
3
3
2
2
4
3
5
2
SOLUTIONS DPP-10
1.
i=
Normal Molar mass
observed molar mass
164
i=
= 2.5
65.6
AB2 → A+2 + 2B–
i = 1 + (n – 1)
2.5 = 1 + (3 – 1)
1.5 = 2
 = 0.75 = 75%
2.
Hg2Cl2 → Hg2+2 + 2Cl–
i = 1 + (n – 1)
i = 1 + (3 – 1) 0.8
i = 1 + 1.6 = 2.6
3.
K3[Fe(CN)6] → 3K+ + [Fe(CN)6]–3
i = 1 + (n – 1) 
i = 1 + (4 – 1)1
i=4
4.
AxBy → xAy+ + yBx–
i = 1 + (n – 1) 
α=
5.
i−1
n−1
=
i−1
x+y−1
Ba(NO3)2 → Ba+2 + 2NO3–
i = 1 + (n – 1) 
2.74 = 1 + (3 – 1) 
2.74 − 1
α=
2
1.74
α=
= 87%
2
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Solutions Part-13
Solutions DPP-11
1.
When equimolar aqueous solutions of glucose, sodium chloride and barium nitrate are compared the
vapour pressure of the solutions will be in the following order: (1) Glucose > NaCl > Ba(NO3)2
(2) Glucose = NaCl = Ba(NO3)2
(3) Ba(NO3)2 > NaCl > Glucose
(4) NaCl > Ba(NO3)2 > Glucose
2.
A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The elevation in boiling point of this
solution is (given Kb = 0.52°C kg mol–1 for H2O)
(1) 0.81
(2) 0.125
(3) 0.48
(4) 1.3
3.
The substance when dissolved in water would decrease the vapour pressure of water to the greatest
extent is:
(1) 0.1 M KCl
(2) 0.1 M urea
(3) 0.1 M BaCl2
(4) 0.1 M NaCl
4.
The vapour pressure will be lowest for
(1) 0.1 M sugar solution
(2) 0.1 M KCl solution
(3) 0.1 M Cu(NO3)2 solution
(4) 0.1 M AgNO3 solution
5.
Which one has the highest boiling point
(1) 0.1 N Na2SO4
(2) 0.1 N MgSO4
(3) 0.1 M Al2(SO4)3
(4) 0.1 M BaSO4
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Solutions Part-13
Question
Answer
Answer Key
1
2
3
1
2
3
4
3
5
3
SOLUTIONS DPP-11
1.
Glucose i = 1
NaCl → Na+ + Cl–, i = 2
Ba(NO3)2 → Ba+2 + 2NO3–,
2.
i=3
Tb = i Kb × m
i = 1 + (n – 1)
HX  H+ + X–
i = 1 + (2 – 1) 0.2 = 1 + 0.2 = 1.2
Tb = 1.2 × 0.52 × 0.2
Tb = 0.125
3.
(1)
KCl → K+ + Cl–
i=2
(2)
Urea
i=1
(3)
BaCl2 → Ba+2 + 2Cl–
i=3
(4)
NaCl → Na+ + Cl–
i=2
4.
Vapour pressure of a solvent is lowered by the presence of solute in it. Lowering in vapour pressure is a
colligative property i.e., it depends on the no. of particles present in the solution. Cu(NO3)2 give the maximum
no. of ions. (i.e., 3) so it causes the greatest lowering in vapour pressure of water.
5.
Al2(SO4)3 gives maximum ion hence it will show highest boiling point.
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Solutions Part-15
Solutions DPP-12
1.
Which of the following solution will have highest freezing point
(1) 1M urea
(2) 1M Na2SO4
(3) 1M NaCl
(4) 1M Al2(SO4)3
2.
A 1.17% solution of NaCl is isotonic with 7.2% solution of glucose calculate the value of i of NaCl
(1) 1
(2) 2
(3) 3
(4) 4
3.
A 0.004M solution of Na2SO4 is isotonic with 0.010M solution of glucose at 25oC temperature. The degree
of dissociation of Na2SO4 is
(1) 25%
(2) 50%
(3) 75%
(4) 85%
4.
What is the freezing point of a solution containing 8.1 g of HBr in 100g water assuming the acid to be
90% ionised (Kf for water = 1.86 K molality–1)
(1) 0.85oC
(2) –3.53oC
(3) 0oC
(4) –0.35oC
5.
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K f for water is1.86 K kg mol–1, the
lowering in freezing point of the solution is:
(1) –0.56 K
(2) –1.12 K
(3) 0.56 K
(4) 1.12 K
6.
Blood is isotonic with
(1) 0.16 M NaCl
(2) Conc. NaCl
(3) 50% NaCl
(4) 30% NaCl
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Solutions Part-15
7.
The osmotic pressure of a solution is given by the relation
(1) P =
RT
C
(2) P =
CT
R
(3) P =
RC
T
(4)
8.
P
= RT
C
What would happen if a thin slice of sugar beet is placed in a concentrated solution of NaCl
(1) Sugar beet will lose water from its cells
(2) Sugar beet will absorb water from solution
(3) Sugar beet will neither absorb nor lose water
(4) Sugar beet will dissolve in solution
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Solutions Part-15
Question
Answer
1
1
2
2
Answer Key
3
4
3
2
5
4
6
1
7
4
8
1
SOLUTIONS DPP-12
1.
(1) Urea,
i=1
(2) Na2SO4,
i=3
(3) NaCl,
i=2
(4) Al2(SO4)3,
i=5
2.
NaCl = glucose
1.17
7.2
i×
× RT =
× RT
58.5
180
7.2 × 58.5 421.2
i=
=
180 × 1.17 210.6
i=2
3.
Na2SO4 = glucose
i × 0.004 × RT = 1 × 0.01RT
0.01
i=
= 2.5
0.004
i = 1 + (n – 1) 
Na2SO4 → 2Na+ + SO42–
2.5 = 1 + (3 – 1) 
1.5
α=
= 0.75 = 75%
2
4.
Tf = i kf × m
HBr  H+ + Br–
i = 1 + (2 – 1) = 1 +  = 1 + 0.9 = 1.9
8.1 1000
ΔTf = 1.9 × 1.86 ×
×
81
100
Tf = 1.9 × 1.86 × 1
Tf = 3.53oC
0 – Tf = 3.53oC
Tf = –3.53oC
5.
Tf = i kf × m
HX  H+ + X–
i = 1 + (2 – 1) 0.2 = 1.2
Tf = 1.2 × 1.86 × 0.5 = 1.12 K
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Solutions Part-15
6.
Blood is isotonic with 0.91%
7.
Osmotic pressure
w
of NaCl solution
v
0.91 g of NaCl is present in 100 mL solution
0.91 1000
M =
×
58.5
100
9.1
=
= 0.156 M
58.5
 0.16 M
P = CRT
P
 = RT
C
8.
Process of osmosis takes place from lower concentrated solution to higher concentrated solution. Thus we can
say that sugar beet will lose water in concentrated NaCl solution.
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Solutions Part-17
Solutions DPP-13
1.
Among the following that forms an ideal solution?
(1) water and methanol
(2) acetone and ethanol
(3) benzene and toluene
(4) water and HCl
2.
On mixing 10 mL of acetone with 40 ml of chloroform the total volume of the solution is
(1) < 50 mL
(2) > 50 mL
(3) = 50 mL
(4) cannot be predicted
3.
The mixture of n-hexane and n-heptane is an example of
(1) ideal solution
(2) non-ideal solution
(3) dilute solution
(4) none
4.
An azeotropic solution of two liquids has boiling point lower than either when it
(1) Shows a negative deviation from Raoult's law
(2) Shows no deviation from Raoult's law
(3) Shows positive deviation from Raoult's law
(4) Is saturated
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Solutions Part-17
Question
Answer
Answer Key
1
2
3
1
3
1
4
3
SOLUTIONS DPP-13
1.
Benzene and Toluene
2.
Acetone and chloroform forms negative deviation so volume is less than 50mL
3.
n hexane and n heptane forms ideal solution
4.
Shows positive deviation from Raoult’s law.
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Solutions Part-18
Solutions DPP-14
1.
The statement “The mass of a gas dissolved in a given mass of a solvent at any temperature is
proportional to the pressure of the gas above the solvent” is
(1) Dalton’s Law of Partial Pressures
(2) Law of Mass Action
(3) Henry’s Law
(4) None of these
2.
Which of the gas will not follow Henry's law?
(1) HCl
(2) He
(3) O2
(4) H2
3.
If solubility of gas ‘X’ is 0.5 gL–1 at 1 bar then its solubility at 3 bar pressure will be
(1) 0.5 gL–1
(2) 1.5 gL–1
(3) 3.0 gL–1
(4) 2 gL–1
4.
Henry's law constant for dissolution of CH 4 in benzene at 298 K is 2 × 105 mm of Hg. Then solubility of
CH4 in benzene at 298 K (in terms of mole fraction) under 760 mm of Hg is:
(1) 1.2 × 10–5
(2) 3.8 × 10–3
(3) 4 × 10–7
(4) 1 × 10–2
5.
Which of the following gas does not obey Henry's law?
(1) NH3
(2) H2
(3) O2
(4) He
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Solutions Part-18
Question
Answer
Answer Key
1
2
3
3
1
2
4
2
5
1
SOLUTIONS DPP-14
1.
Henry’s Law
2.
HCl
3.
Solubility ×Partial pressure
0.5 1
=
x
3
x = 1.591 L
4.
P = KH × Xgas
760 = 2 × 105 × Xgas
760
Xgas =
= 3.8 × 10–3
2×105
5.
Those gases which can react with solvent do not follow henry’s Law i.e. NH3
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