The vapour pressure of benzene at 90 ° C is 1020 torr. A solution of 15 g of a solute in 58.5 g benzene has a vapour pressure of 990 torr. The molecular weight of the solute is

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Using Raoult's law for relative lowering of vapour pressure as follows
$\frac{P_{o} - P_{s}}{P_{o}} = \frac{w \times M}{m \times W}$
Here $P_{o} = V P$ or pure solvent, $P_{s} =$ VP of solution
or $\frac{1020 - 990}{1020} = \frac{5 \times 78}{m \times 58.5}$
So $m = 226.6$ .

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