Question

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

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, m = mol. wt. of solute

W = mass of solvent, M = mol. wt. of solvent

$\Rightarrow \frac{121.8-120.2}{121.8}=\frac{\frac{15}{m}}{\frac{250}{78}}$

$\Rightarrow m=356.7g$