The vapour pressure of pure liquid solvent `A` is `0.80 atm`. When a non-volatile substance `B` is added to the solvent, its vapour pressure drops to `0.60 atm`, the mole fraction of component `B` in the solution is

Options

A. 0.25
B. `0.50`
C. 0.75
D. `0.90`

Answer

A

Explanation

Lowering of vapour pressure depends on the sum of the mole fraction of different solutes.
i.e. `(P^(0)-P)/(P^(0))=chi_(2)`
`(0.80-0.60)/(0.80)=chi_(2)`
`chi_(2)=(1)/(4)=0.25`.