The vapour pressures of ethanol and methanol are 42.0 mm and 88.5 mmHg respectively. An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. The mole fraction of methanol in the vapour is :

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0.502

0.513

0.556

0.467

answer is C.

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Detailed Solution

$P_{m} = P_{MeOH}^{o} X_{MeOH}^{o} + P_{EtOH}^{o} X_{EtOH}$

Thus $P_{M} = 88 .5 \times [\frac{\frac{16}{32}}{\frac{16}{32} + \frac{46}{46}}] + 42 \times [\frac{\frac{46}{46}}{\frac{16}{32} + \frac{46}{46}}] = 57 .5$

Now $P'_{MeOH} = P_{MeOH}^{o} . X_{MeOH (l)} = P_{m} \times X_{MeOH (g)}$

$\begin{array}{l} ∴ 88 .5 \times [\frac{\frac{16}{32}}{\frac{16}{32} + \frac{46}{46}}] = 57 .5 X_{MeOH (g)} \\ ∴ X_{MeOH} = 0 .513 \end{array}$

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