A solution of a non-volatile solute in water freezes at $-0{.40}^{\mathrm{o}}\mathrm{C}$. The vapour pressure of pure water at 298 K is 23.51 torr. For water, ${\mathrm{K}}_{\mathrm{f}}=1.86 \mathrm{K} {\mathrm{mol}}^{-1} \mathrm{kg}$. Thus, vapour pressure of the solution (in torr) 

For $\left(\frac{{\mathrm{w}}_1}{{\mathrm{m}}_1}\right)$ mole of non-volatile solute in $w_2$ g solvent, depression in freezing point is 

$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{f}}=\frac{1000{\mathrm{K}}_{\mathrm{f}}{\mathrm{w}}_1}{{\mathrm{m}}_1{\mathrm{w}}_2}\\ 0.40=\frac{1000\times 1.86{\mathrm{w}}_1}{{\mathrm{m}}_1{\mathrm{w}}_2}\\ \therefore \frac{{\mathrm{w}}_1}{{\mathrm{m}}_1{\mathrm{w}}_2}=\frac{40}{1000\times 186}\end{array}$

By Raoult's law, 

$\frac{{\mathrm{p}}^{\mathrm{o}}-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^{\mathrm{o}}}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}=\frac{{\mathrm{w}}_1}{{\mathrm{m}}_1}\frac{{\mathrm{m}}_2}{{\mathrm{w}}_2}$

$1-\frac{{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^{\mathrm{o}}}=\frac{40\times 18}{1000\times 186}=\frac{72}{18600}$

$\begin{array}{l}1-\frac{{\mathrm{p}}_{\mathrm{s}}}{23.51}=3.87\times {10}^{-3}\\ {\mathrm{p}}_{\mathrm{s}}=19.64\text{ torr}\end{array}$