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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:

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