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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

0.6

B

0.5

C

0.2

D

0.8

Text Solution

Verified by Experts

The correct Answer is:
D
`p_(A) = x_(A) xx p_(A)^(@) " " or " " 32 = x_(A) xx 40`
or `x_(A) = (32)/(40) = 0.8`
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