A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mmHg at 300 K. The vapour pressure of propyl alcohol is 200 mm $Hg$ . If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm $Hg$ ) at the same temperature will be

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700

360

350

300

answer is A.

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Detailed Solution

Using the expression $p = x_{A} p_{A} + x_{B} p_{B};$ (A is ethyl alcohol and B is propyl alcohol)

we get $290 mmHg = 0.6 p_{A} + 0.4 \times 200 mmHg$

This give $p_{A} = \frac{(290 - 0.4 \times 200) mmHg}{0.6} = 350 mmHg$

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