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Methanol (p* = 90 mmHg at 300 K) and ethanol (p* = 51 mmHg at 300 K) form very nearly an ideal solution. The total vapour pressure of a solution obtained by mixing 23 g ethanol with 32 g methanol would be

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detailed solution

Correct option is C

Amount of methanol, $n_{1} = \frac{m_{1}}{M_{1}} = \frac{32 g}{32 g {mol}^{- 1}} = 1.0 mol$

Amount of ethanol, $n_{2} = \frac{m_{2}}{M_{2}} = \frac{23 g}{46 g {mol}^{- 1}} = 0.5 mol$

Mole fraction of methanol, $x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = \frac{1.0 mol}{(1.0 + 0.5) mol} = 2 / 3$

Total Vapour pressure, $p = x_{1} p_{1}^{*} + x_{2} p_{2}^{*} = [(2 / 3) (90) + (1 / 3) (51)] mmHg = 77 mmH g$

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