The vapour pressure of two pure liquids A and B that forms an ideal solution are 300 and 800 torr respectively at temperature T. A mixture of the vapours of A and B (for which the mole fraction of A is 0.25) is slowly compressed at temperature T. The vapour pressure of this condensate on 100% condensation of vapoure is measured to be ‘P’. What is the value of 684 –P?

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answer is 9.

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Detailed Solution

$\begin{array}{l} \frac{X_{A}^{'}}{X_{B}^{'}} = \frac{p_{A}^{0} X_{A}}{p_{B}^{0} X_{B}} \Rightarrow \frac{X_{A}}{X_{B}} = \frac{X_{A}^{'}}{X_{B}^{'}} \times \frac{p_{A}^{0}}{p_{B}^{0}} = \frac{0 .25}{0 .75} \times \frac{800}{300} = \frac{8}{9} \\ \Rightarrow X_{B} = 0 .53 \\ \Rightarrow P = 300 \times 0 .47 + 800 \times 0 .53 = 565 torr \\ So, 574 - P = 9 \end{array}$

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