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A solution contains 1 mol of pentane $(p^{} = 450 mmHg)$ and 4 mol of hexane $(p^{} = 150 mmHg)$ , the mole fraction of pentane in vapour phase will be

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0.429

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detailed solution

Correct option is B

In vapour phase, we will have

$\begin{array}{l} p_{pentane} = {1 / (1 + 4)} (450 mmHg) = 90 mmHg; p_{pentane} = {4 / (1 + 4)} (150 mmHg) = 120 mmHg \\ y_{pentane} = \frac{p_{pentane}}{p_{pentane} + p_{hexane}} = \frac{90}{90 + 120} = 0.429 \end{array}$

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A solution contains 1 mol of pentane p∗=450mmHg and 4 mol of hexane p∗=150mmHg, the mole fraction of pentane in vapour phase will be