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Question

At 300 K, two pure liquids A and B have vapour pressures 150 mm Hg and 100 mm Hg respectively. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is:

A
0.6
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B
0.5
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C
0.8
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D
0.4
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Solution

The correct option is D 0.4
Let the moles of A=x

The moles of B=x[ mixture is equimolar.]

Mole fraction of A=xx+x=0.5 mole

= mole fraction of B

Thus, ρT=ρ0AχA+ρ0BχB

ρT=150×0.5+100×0.5=125

Mole fraction of B in the vapour mixture =ρ0BχBρT

=100×0.5125=0.4.

Hence, option D is correct.

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