What is the vapour pressure of solution prepared by mixing $25.5 \mathrm{g} \mathrm{of} {\mathrm{CHCl}}_3\left({\mathrm{p}}^0=200 \mathrm{mm}\right) \mathrm{and} 40 \mathrm{g} \mathrm{of} {\mathrm{CH}}_2{\mathrm{Cl}}_2 \left({\mathrm{p}}^0=415 \mathrm{mm}\right)\mathrm{at} 298 \mathrm{K}$? 

$\begin{array}{l}\text{ Molar mass of }{\mathrm{CH}}_2{\mathrm{Cl}}_2=12\times 1+1\times 2+35.5\times 2\\ =85\mathrm{g}{\mathrm{mol}}^{-1}\end{array}$
$\begin{array}{l}\text{ Molar mass of }{\mathrm{CHCl}}_3=12\times 1+1\times 1+35.5\times 3\\ =119.5\mathrm{g}{\mathrm{mol}}^{-1}\end{array}$
            $\text{ Moles of }{\mathrm{CHCl}}_3=\frac{25.5}{119.5}=0.213\mathrm{mol}$
          $\text{ Moles of }{\mathrm{CH}}_2{\mathrm{Cl}}_2=\frac{40\mathrm{g}}{85\mathrm{g}{\mathrm{mol}}^{-1}}=0.47\mathrm{mol}$
$\text{ Total number of moles }=0.47+0.213=0.683\mathrm{mol}$
                           $\begin{array}{l}{\mathrm{x}}_{{\mathrm{CH}}_2{\mathrm{Cl}}_2}=\frac{0.47\mathrm{mol}}{0.683\mathrm{mol}}=0.688\\ {\mathrm{x}}_{{\mathrm{CHCl}}_3}=1.00-0.688=0.312\end{array}$
Using equation,   
$\begin{array}{l}{\mathrm{p}}_{\text{total }}={\mathrm{p}}_1^{\circ }+\left({\mathrm{p}}_2^{\circ }-{\mathrm{p}}_1^{\circ }\right){\mathrm{x}}_2\\ =200+(415-200)\times 0.688\\ =200+147.9\\ =347.9\mathrm{mmHg}\end{array}$