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Q.

Vapour pressure of benzene at $30 ° C$ is 121.8 mm Hg. When 15g of a non volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm Hg. The molecular weight of the solute (Considered as a dilute solution) [Nearest Integer]
(Mo. Wt. of solvent = 78)

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answer is 356.

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Detailed Solution

From Raoult’s law
$\frac{P ° - P}{P °} = \frac{w}{m} \times \frac{M}{W}$

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

Or $m = \frac{15 \times 78}{250} \times \frac{121.8}{1.6} = 356.2$

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