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NCERT Class XII Chemistry
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Question : 17 of 53
Marks: +1, -0
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,P^{0}=12.3\,\text{kPa},\ P_{s}=?\ \text{m}=1, No. of moles of solute =1=1,
No. of moles of water =100018=55.5=\frac{1000}{18}=55.5
∴  \therefore\; Mole fraction of solute =11+55.5=0.0177=\frac{1}{1+55.5}=0.0177
From Raoult's law, P0−PsP0=x2\frac{P^{0}-P_{s}}{P^{0}}=x_{2} ,
i.e., 12.3−Ps12.3=0.0177\frac{12.3-P_{s}}{12.3}=0.0177 or Ps=12.08 kPaP_{s}=12.08\,\text{kPa}
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