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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)=X_(A)P_(A)^(@)`
`32=X_(A)40`
`therefore X_(A)=(32)/(40)=0.8`.
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