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Question

At 80 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 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,
PT=PoAxA+PoBxB
xA and xB mole fraction of A and B
760=520xA+(1000(1xA)
xA=12=0.5
So, 50 mol % of A is present in the mixture.

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