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At $25^{0} C$ vapour pressure of volatile liquid ‘A’ in pure state is 60 torr. ‘X’ moles of ‘B’ volatile liquid is added to 4 moles of ‘A’ at $25^{0} C$ .The vapour pressure of resulting solution was found to be 72 torr, and mole fraction of ‘A’ in vapour phase is $\frac{1}{3}$ , the value of ‘X’ is

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answer is 6.

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Detailed Solution

$n_{A} = 4 n_{B} = X X_{A} = \frac{4}{(4 + X)} X_{B} = \frac{X}{4 + X} y_{A} = \frac{X_{A} . P_{A} 0}{P} \frac{1}{3} = \frac{60 X_{A}}{P} \frac{24}{60} = X_{A} \frac{24}{60} = \frac{4}{4 + X} X = 6 m o l e s$

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