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:

Options

A. 0.5
B. 0.6
C. 0.7
D. 0.8

Answer

D

Explanation

`"Given, vapour pressure of pure"A=40mm" Hg"`
`"Vapour pressure of A in solution" =32mm" Hg"`
`"According to Raoult's law",P_(A)=P_(A)^(@)x_(A)`
`" or "x_(4)=(P_(A))/(P_(A)^(@))=(32mm" Hg")/(40mm" Hg")=0.8`