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NCERT Class XII Chemistry
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Question : 50
Total: 53
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:  
Given, P0=23.8 mm Hg
w2=50 g,M2( urea )=60 g mol−1,
Ps=?
P0−P s
P0
=?
w1=850 g,M1( water )=18 g mol−1
∴‌n2=
50
60
=0.83
∴n1=
850
18
=47.22
Applying Raoult's law,
P0−Ps
P0
=
n2
n1+n2
or,
P0−Ps
P0
=
0.83
47.22+0.83

=
0.83
48.05
=0.017
Thus, relative lowering of vapour pressure = 0.017
Again,
ΔP
P0
=0.017
⇒ΔP=0.017×23.8
or, P0−Ps=0.017×23.8
or, Ps=23.8−0.017×23.8
or Ps=23.4 mm Hg
Thus, vapour pressure of water in the solution = 23.4 mm Hg
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