The vapour pressure of water is 12.3 k `P_(a)` at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it.

Explanation

Given molality of solution = 1m
Vapour pressure of water `P_(0)=12.3" kP"_(a)`
Number of moles of solute `=1.(n_(s))`
Number of moles of water `=(1000)/(18)`
`=55.55(n_(0))`
Mole fraction of solute `(X_(s))=(n_(s))/(n_(0)+n_(s))`
`=(1)/(55.55+1)`
`=(1)/(56.55)=0.0177`
Mole fraction of water `(X_(0))=1-X_(s)`
`=1-0.0177=0.9823`
Vapour pressure of solution `(P_(s))=P_(0)X_(0)`
`=12.3xx0.9823=12.08" kP"_(a)`