Question

Vapour pressure of benzene at $30^{\circ} C$ is $121.8\, mm$. when $15\, g$ of a non-volatile solute is dissolved in $250\, g$ of benzene, its vapour pressure is decreased to $120.2\, mm$. The molecular weight of the solute is

• A. 35.67 g
• B. 356.7 g
• C. 432.8 g
• D. 502.7 g

Answer 1

Answer: (B)

356.7 g

Answer 2

As, $\frac{P^{0}-P}{P^{0}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}$ $P^{0}$ = vapour pressure of pure component P = vapour pressure in solution w = mass of solute, $\quad$ m = mol. wt. of solute W = mass of solvent, M = mol. wt. of solvent $\Rightarrow \frac{121.8-120.2}{121.8}=\frac{\frac{15}{m}}{\frac{250}{78}}$ $\Rightarrow m = 356.7 \,g$