The vapour pressure of benzene at $80^{\circ} C$ is 1020 torr. 5 g of a solute is dissolved in 58.5

Question

The vapour pressure of benzene at $80^{\circ} C$ is 1020 torr. 5 g of a solute is dissolved in 58.5 g of benzene and the resulting solution has a vapour pressure of 990 torr. Molecular mass of the solute is

Professor PC Thomas Classes · Answer

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}+n_{2}} \approx \frac{n_{2}}{n_{1}}=\frac{w_{2} M_{1}}{M_{2} w_{1}} \frac{1020-990}{1020}=\frac{5 	imes 78}{M_{2} 	imes 58.5} \frac{30}{1020}=\frac{5 	imes 4}{M_{2} 	imes 3} \quad M_{2}=226$