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Solutions (Numericals)
1) The solubility of N2 gas in water at 250C and 1 bar is 6.85×10−4mol/L.
Calculate a) Henry’s law constant b) molarity of N2 gas dissolved in
water under atmospheric conditions when partial pressure of N2 in
atmosphere is 0.75 bar.
−4 3
S 6.85 ×10 mol /dm
Ans. – a) KH = =
P 1 ¯¿ ¿

= 6.85 ×10−4 mol /L /¯¿

b) S = KH P = ¿ ¿
= 5.138 ×10−4 mol /L
2) The Henry’s law constant of methyl bromide (CH3Br) is 0.159
mol/L/bar at 250C. What is the solubility of methyl bromide in water
at 250C and at pressure of 130 mm Hg.
1
Ans. – Since, P = 130 mm Hg × 760 mm Hg/atom

= 0.171 atm ×1.013 ¯¿ atom

= 0.173 bar
Here, KH = 0.159 mol/L/bar
According to Henry’s law, S = KH P
=¿
= 0.271 M .
3) Henry’s constant for CH3Br (g) is 0.159 mol dm-3bar-1 at 250C.
Calculate its solubility in water at 250C, if its partial pressure is 0.164
bar.
Ans. – Given – KH = 0.159 mol dm-3bar-1,
P = 0.164 bar
Formula : S = KH P
= 0.159 × 0.164
= 0.02 mol dm-3.
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4) Henry’s constant for solubility of methane in benzene is 3.25×10−2

mol L-1 atm-1 at constant temperature. Calculate solubility of methane
at 1 bar pressure and at same temperature.
Ans. - Given – KH = 3.25×10−2 mol L-1 atm-1,
P = 1 bar
−2
3.25× 10
KH = 3.25×10 mol L atm
−2 -1 -1
= = 3.21×10−2 mol L-1 bar-1
1.013

S = KH P
= 3.21×10−2 mol L-1 bar-1 × 1 bar
= 3.21 mol L-1.
5) For a gas, the Henry’s constant is 1.25×10−3 mol dm-3 atm-1 at 250C.
Calculate the solubility of the given gas at 2.5 atm and 250C.
Ans. – Given – KH = 1.25×10−3 mol dm-3 atm-1,
P = 2.5 atm,
S=?
By using Henry’s law, S = KH P
= 1.25×10−3 mol dm-3 atm-1×2.5 atm
= 3.125×10−3 mol dm-3
6) The solubility of dissolved oxygen to 170C is 2.6×10−3 mol dm-3 at 2
atm. Find its solubility at 8.4 atm and 270C.
Ans. – Solubility of O2 = S1 = 2.6×10−3 mol dm-3,
Solubility of O2 = S2= ?
Initial pressure of O2 = P1 = 2 atm,
Final pressure of O2 = P2 = 8.4 atm
(i) By Henry’s law S1 = KH P1
S1
KH = P
1
−3
2.6 ×10
¿
2
= 1.3 ×10−3 mol dm-3 atm-1
(ii) By Henry’s law S2 = KH P2
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= 1.3 ×10−3 × 8.4

= 10.92 ×10−3
¿ 1.092 ×10 mol dm-3
−2

Solubility of O2 = 1.092 ×10−2 mol dm-3

7) Henry’s law constant for the solubility of methane in benzene is
4.27 ×10−5 mm-1 Hg mol dm-3 at constant temperature. Calculate the
solubility of methane at 7600 mm Hg pressure at same temperature.
Ans. – Given Henry’s law constant, KH = 4.27 ×10−5 mm-1 Hg mol dm-3,
Pressure of the gas = P = 760 mm Hg
KH = 4.27 ×10−5 ×760
= 3245×10−5 atm-1 mol dm-3
= 3.245 ×10−2 atm-1 mol dm-3
760
P = 760 mm Hg = 760 atm = 1 atm

By Henry’s law, S = KH P
= 3.245 ×10−2 atm-1 mol dm-3× 1 atm
= 3.245 ×10−2 mol dm-3
8) The solubility of ethane at 250C is 0.92×10−3 g dm-3 at 1000 mm Hg
pressure. Calculate Henry’s law constant.
Ans. – Solubility of ethane = S = 0.92×10−3 g dm-3,
Pressure of ethane = P = 1000 mm Hg
Molar mass of ethane (C2H6) = 30 g/mol,
Henry’s law constant, KH= ?
S = 0.92×10−3 g dm-3
0.92 −3
= 30 ×10
−5
¿ 0.3067 ×10 mol dm-3
1000
P = 1000 mm Hg = 760 atm = 1.316 atm.

S
By Henry’s law, KH = P
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−5
3.067 ×10
=
1.316
= 2.22 ×10−5 mol dm-3 atm-1
9) The solubility of nitrogen at 300C is 2.5 ×10−3 g dm-3 at 760 m
pressure. What will be its solubility in mol dm-3 at 20, 000 mm and
same temperature.
Ans. – Initial solubility of N2 = S1 = 2.5 ×10−3 g dm-3,
Initial pressure = P1 = 760 mm
Final pressure = P2 = 20, 000 mm,
Final solubility = S2= ?
Molar mass of N2 = 28 g/mol
S1 = 2.5×10−3 g dm-3
2.5 −3
= 28 ×10

¿ 8.93 ×10
−5
mol dm-3
760
P1 = 760 atm = 1 atm

20000
P2 = 760 atm = 26.32 atm

By Henry’s law, S1 = KH P1
S1
KH = P
1

−5
8.93 ×10
=
1
= 8.93 ×10−5 mol dm-3 atm-1
S2 = K H P 2
= 8.93 ×10−5 × 26.32
= 2.35 ×10−3 mol dm-3
Alternative method –
S1 = KH P1 and S2 = KH P2
S2 K H P2 P2
Since, S = K P = P
1 H 1 1
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P2
Hence, S2 = S1× P
1

−5 20 , 000
¿ 8.93 ×10 ×
760

= 2.35 ×10−3 mol dm-3

10) Fish generally needs O2 concentration in water at least 3.8 mg/L
for survival. What partial pressure of O2 above the water is needed for
the survival of fish? Given the solubility of O2 in water at O0C and 1 atm
partial pressure is 2.2×10−3mol/L
S
Ans. – KH = P
−3
2.2× 10 mol /L
=
1.013 ¯¿ ¿
= 2.17 ×10−3 mol L-1 bar-1
−3
3.8× 10
Required concentration of O2 = 3.8 mg/L = = 1.18×10−4 mol/L
32
S
Pressure needed = P = K
H

−4
1.18× 10
¿ −3
2.17 ×10

= 0.054377 atm.
11) The vapour pressures of pure liquids A and B are 450 mm Hg and
700 mm Hg respectively at 350 K. Find the composition of liquid and
vapour if total vapour pressure is 600 mm.
Ans. – i) Composition of A and B in the solution are x1 and x2
Here P1o = 450 mm Hg,
P20 = 700 mm Hg and
P = 600 mm Hg
Hence, 600 mm Hg = (700 mm Hg – 450 mm Hg) x2 + 450 mm Hg
= 250 x2 + 450
600 – 450 = 250 x2
150 = 250 x2
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Or x2 = 250/150 = 0.6
Since, x1 = 1 – x2
= 1 – 0.6 = 0.4
ii) Composition of A and B in vapour are y1 and y2 respectively.
P1 = y1 P and P2 = y2 P
Then according to Raoult’s lawP1 = P10 x1 and P2 = P20 x2
P1
Hence, y1 =
P
0
P x
= 1 1
P
450 mm Hg× 0.4
= 600 mm Hg

= 0.3
y2 = 1 – y1
= 1 – 0.3
= 0.7
12) An aqueous solution is made by dissolving 20 g of non volatile
solute (molar mass = 60 g/mol) in 90 g of water at 303 K. If the vapour
pressure of pure water at 303 K is 32.8 mm Hg, what would be the
vapour pressure of the solution.
Ans. – Given :Vapour pressure of pure water = P10 = 32.8 mm Hg
Mass of non volatile solute = 20 g,
Molar mass of non volatile solute = 60 g/mol
Vapour pressure of solution (P1) = ?
90
Number of moles of water = 18 = 5 mol

20
Number of moles of non volatile solute = 60 = 0.333 mol

5
Mole fraction of water = x1 = 5+0.333 = 0.938

Vapour pressure of solution, P1 =P10 x1

= 32.8 × 0.938
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= 30.77 mm Hg
13) A solution is prepared by dissolving 394 g of a non-volatile solute
in 622 g of water. The vapour pressure of solution is found to be 30.74
mm Hg at 30 0C. If vapour pressure of water at 30 0C is 31.8 mm Hg,
what is the molar mass of solute?
0
P 1−P1 W2M1
Ans. - 0 =M W
P 1 2 1

Since, W2 = 394 g,
W1 = 622 g,
M1 = 18 g/mol,
P1 = 30.74 mm Hg,
P0= 31.8 mm Hg.
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Substitute these values in above equation, we get

31.8 mm Hg−30.74 mm Hg 394 g ×18 g mol−1
31.8 mm Hg
= M 2 × 622 g
−1
11.4 g mol
0.033 = M2
−1
11.4 g mol
Hence, M2 = = 342 g/mol
0.033

14) The vapour pressure of pure benzene (molar mass 78 g/mol) at a

certain temperature is 640 mm Hg. A non-volatile solute of mass
2.315 g is added to 40 g of benzene. The vapour pressure of the
solution is 600 mm Hg. What is the molar mass of solute?
Ans. - P10= 640 mm Hg,
P1 = 600 mm Hg,
W1 = 49 g,
W2 = 2.315 g
0
P 1−P1 W2 M1
0 =
P 1
M2 W 1

640 mm Hg−600 mm Hg 2.315 g−78 g/mol

Then, 640 mm Hg
= M 2 × 40 g
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2.315 g−78 g/mol × 640 mm Hg

M2 = 40 mm Hg × 40 g

= 72.23 g mol
15) The vapour pressure of water at 200C is 17 mm Hg. What is the
vapour pressure of solution containing 2.8 g urea in 50 g of water?
Ans. – Given : Vapour pressure of pure water = P10 = 17 mm Hg,
Mass of urea (W2) = 2.8 g
Molar mass of water (W1) = 50 g,
Vapour pressure of the solution (P1) = ?
Molar mass of urea (NH2CONH2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g/mol
Molar mass of water = 18 g/mol
0
P 1−P1 W2 M1
0 =
P 1
M2 W 1

17−P1 2.8× 18
=
17 50× 60
17−P1
= 0.0168
17

17 – P1 = 0.0168 × 17
= 0.2856
P1 = 16.71 mm Hg
16) The normal boiling point of ethyl acetate is 77.06 0C. A solution of
50 g of a non-volatile solute in 150 g of ethyl acetate boils at 84.27 0C.
Evaluate the molar mass of solute if Kb for ethyl acetate is 2.77 0C
kg/mol.
1000 K b W 2
Ans. - M2 = ∆ T W
b 1

Since, W2 = 50 g,
W1 = 150 g
△Tb = Tb - T b
0

= 84.27 0C – 77.06 0C = 7.21 0C = 7.21 K

Kb = 2.77 0C kg/mol = 2.77 K kg/mol
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Substitute these values in above equation we get,

1000 g /Kg ×2.77 K Kg/mol × 50 g
M2 = 7.21 K ×150 g
= 128 g/mol

17) 3.795 g of sulphur is dissolved in 100 g of carbon disulfide. This

solution boils at 319.81 K. What is the molecular formula of sulphur in
solution? The boiling point of the solvent is 319.45 K.
(Given that Kb for CS2 = 2.42 H Kg/mol and atomic mass of S = 32u)
1000 K b W 2
Ans. - M2 = ∆ T W
b 1

Since, W1 = 100 g, W2 = 3.795 g

△Tb = Tb - T 0b
= 319.81K – 319.45 K
= 0.36 K
g kg
1000 × 2.42 K × 3.795 g
M2 = Kg mol
0.36 K × 100 g

= 255.10 g/mol
Atomic mass of S = 32 u
molar mass of S
Therefore, number of atoms in a molecule of S = atomic mass of S

2.55.1
= 32

= 7.92
=8
Hence, molecular formula is S8 in CS2.
18) The freezing point of pure benzene is 278.4 K. calculate the
freezing point of the solution when 4 g of a solute having molecular
weight 100 g/mol is added to 200 g of benzene (Kf for benzene = 5.12
K kg/mol)
Ans. – Given :T 0f =278.4 K ,
W2 = 4 g,
M2 = 100 g/mol,
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W1 = 200 g, Kf 5.12 K kg/mol,

Tf = ?
1000 K f W 2
M2 = ∆ T W
f 1

Hence, △Tf¿
1000 K f W 2
M 2W 1
1000× 5.12× 4
= 100 ×200

= 1.024 K
0
Now, ,△Tf¿ T f −T f
= 278.4 K – 1.024 K
= 277.376 K
Hence, freezing point of solution is 277.376 K