The vapour pressures of pure .liquids A and B are $400$ and $600 m m H g$ , respectively at $298 K$ . On mixing the two liquids,
the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is $0.5$ in the mixture.
The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase respectively are

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$500 m m H g, 0.5, 0.5$

$450 m m H g, 0.4, 0.6$

$450 m m H g, 0.5, 0.5$

$500 m m H g, 0.4, 0.6$

answer is D.

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Detailed Solution

$P_{totol} = X_{A} P_{A}^{0} + X_{B} P_{B}^{0}$

$= 0.5 \times 400 + 0.5 \times 600 = 500 mmHg$

Now, mole fraction of A in vapour,

$Y_{A} = \frac{P_{A}}{P_{total}} = \frac{0.5 \times 400}{500} = 0.4$

and mole fraction of $B$ in vapour

$Y_{B} = 1 - 0.4 = 0.6$

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