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Q.

The vapour pressure of two volatile liquids A and B at $25^{0} C$ are 50 torr and 100 torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is $\frac{x}{17} .$ The value of ‘x’ is……..

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answer is 14.

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Detailed Solution

$P_{A}^{0} = 50 t o r r P_{B}^{0} = 100 t o r r;$
$X_{A} = 0.3 X_{B} = 0.7$
$P_{t o t} = P_{A}^{0} \times A + P_{B}^{0} \times B = 50 \times 0.3 + 100 \times 0.7$
$= 15 + 70 = 85$
$y_{B} =$ mole fraction of ‘B’ in vapour phase)
$= \frac{P_{B}}{P_{t o t}} = \frac{P_{B}^{0} \times B}{P_{t o t}} = \frac{70}{85} = \frac{14}{17};$
$\frac{x}{17} = \frac{14}{17}; x = 14$

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The vapour pressure of two volatile liquids A and B at 250C are 50 torr and 100 torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is x17.The value of ‘x’ is……..