$\text{ Vapour pressure of benzene at }{30}^{\circ }\mathrm{C}\text{ is }121.8\mathrm{mmHg}\text{. When }15\mathrm{g}\text{ of a non volatile solute is dissolved in }250\mathrm{g}\text{ of benzene its vapour pressure decreased}$
$\mathrm{to} 120.2\mathrm{mmHg}\text{. The molecular weight of the solute }(\mathrm{Mo}\text{. wt. of solvent }=78)$

$$\begin{gathered} \text{ Given vapour pressure of pure solvent } \\ \left({\mathrm{P}}^{\circ }\right)=121.8\mathrm{mmHg};\text{ Weight of solute }\left(\mathrm{w}\right)=15\mathrm{g} \\ \text{ Weight of solvent }\left(\mathrm{W}\right)=250\mathrm{g}\text{; Vapour pressure of solution } \\ \left(\mathrm{P}\right)=120.2\mathrm{mmHg}\text{ and Molecular weight of solvent }\left(\mathrm{M}\right)=78 \\ \text{ From Raoult's law }=\frac{{\mathrm{P}}^{\mathrm{o}}-\mathrm{P}}{{\mathrm{P}}^{\mathrm{o}}}=\frac{\mathrm{w}}{\mathrm{m}}\times \frac{\mathrm{M}}{\mathrm{W}} \\ \frac{121.8-120.2}{121.8}=\frac{15}{\mathrm{m}}\times \frac{78}{250} \\ \text{ or }\mathrm{m}=\frac{15\times 78}{250}\times \frac{121.8}{1.6}=356.2 \end{gathered}$$