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Vapour Pressure

At 80∘ C, the...

Question

At $80^{\circ} C$ , the vapour pressure of a pure liquid $A$ is $520 mm Hg$ and that of a pure liquid of $B$ is $1000 mm Hg$ . If the ideal mixture of liquids $A$ and $B$ boils at $80^{\circ} C$ and 1 atm pressure, the amount of solution A in the mixture is: (1 atm = 760 mm Hg)

A

50 mol %

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B

52 mol %

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C

34 mol %

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D

48 mol %

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Solution

The correct option is A $50 mol %$
We know that,
$P_{T} = P_{A}^{o} x_{A} + P_{B}^{o} x_{B}$
$x_{A} a n d x_{B}$ mole fraction of A and B
$760 = 520 x_{A} + (1000 (1 - x_{A})$
$x_{A} = \frac{1}{2} = 0.5$
So, 50 mol % of A is present in the mixture.

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