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The vapour pressure of a pure liquid A i...

The vapour pressure of a pure liquid `A` is `40 mm Hg` at `310 K`. The vapour pressure of this liquid in a solution with liquid `B` is `32 mm Hg`. The mole fraction of `A` in the solution, if it obeys Raoult's law, is:

A

0.8

B

0.5

C

0.2

D

0.4

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(A)=40 mm Hg`,`P_(A)=P_(A)^(@)chi_(A)=32 rArr chi_(A) = 0.8`,`chi_(B)=0.2`
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