At $50 ° C$ , the vapour pressure of pure benzene is 268 torr. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of 167 torr at the same temperature is
$(molar mass of benzene = 78 g {mol}^{- 1})$

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0.605

0.405

0.505

0.705

answer is C.

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Detailed Solution

$p_{benzene} = p_{benzene}^{0} x_{benzene} 167 = 268 x_{benzene} x_{benzene} = \frac{167}{268} = 0.62 x_{solute} = 0.377 (1 - 0.62) \frac{x_{solute}}{x_{benzene}} = \frac{0.377}{0.62} = 0.605$

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