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At 300 K the vapour pressure of two pure...

At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of a and B, the vapoure is 300 mm Hg, the mole fractions of a in the vapour phase, respectively, are -

A

1/2 and 1/10

B

1/4 and 1/6

C

1/4 and 1/10

D

1/2 and 1/6

Text Solution

Verified by Experts

The correct Answer is:
D
`Y_(A)=(P_(A)^(@)X_(A))/(P_(A)^(@)X_(A)+ P_(B)^(@)X_(B))=(100. 1/2)/(100. 1/2+500. 1/2)`
`{:(=50/(50+250),,|,:' Ps=(p_(A)^(@)-p_(B)^(@))X_(A)+p_(B)^(@),),(=50/300,,300=(-400)X_(A)+500,,),(=1//6,,X_(A)=1/2,,),(,,:. X_(B)=1/2,,):}`
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