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NCERT Class XII Chemistry
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Question : 19
Total: 53
A solution containing 30 g of non-volatile solute exactly in 90 g water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to the solution, the new vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K.
Solution:  
(i) Suppose the molar mass of the solute =M g mol−1
No. of moles of solute, n2=
30
M
 moles
No. of moles of solvent, H2O,n1=
90 g
18 g mol−1
=5 ‌moles
From Raoult's law,
P0−Ps
P0
=
n2
n1+n2
, i.e.,
P0−2.8
P0
=
30
M
5+
30
M

or, 1−
2.8
P0
=
30
M
5+
30
M

or
2.8
P0
=1−
30
M
5+
30
M
 =
5+
30
M
−
30
M
5+
30
M
 =
5
5+
30
M

or,
P0
2.8
=
5+
30
M
5
=1+
6
M
‌‌‌‌‌‌‌‌....(i)
After adding 18 g of water, n1 = 6 moles
P0−2.9
P0
=
30
M
6+
30
M
‌‌ or, 1−
2.9
P0
=
30
M
6+
30
M

or,
2.9
P0
=1−
30
M
6+
30
M
 =
6+
30
M
−
30
M
6+
30
M
 =
6
6+
30
M

or
P0
2.9
=
6+
30
M
6
=1+
5
M
‌‌‌‌‌‌‌‌‌‌‌‌‌‌....(ii)
Dividing eqn. (i) by eqn. (ii), we get
2.9
2.8
=
1+
6
M
1+
5
M
 or, 2.9 (1+
5
M
 )=2.8 (1+
6
M
)
or, 2.9+
14.5
M
=2.8+
16.8
M
 or,
2.3
M
=0.1 or M=23 u
(ii) Putting M = 23 in eqn. (i) we get
P0
2.8
=1+
6
23
=
29
23
 or P0=
29
23
×2.8=3.53 ‌kPa
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