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NCERT Class XII Chemistry
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Question : 17
Total: 53
The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a solute in it.
Solution:  
Given P0=12.3 ‌kPa,Ps=? m=1, No. of moles of solute =1,
No. of moles of water =
1000
18
=55.5
∴‌ Mole fraction of solute =
1
1+55.5
=0.0177
From Raoult's law,
P0−Ps
P0
=x2 ,
i.e.,
12.3−Ps
12.3
=0.0177 or Ps=12.08 ‌kPa
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