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The vapour pressure of pure liquid is 1....

The vapour pressure of pure liquid is 1.2 atm. When a non-volatile substance B is mixed in A, then its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is

A

`0.15`

B

`0.25`

C

`0.50`

D

`0.75`

Text Solution

Verified by Experts

The correct Answer is:
C
According to Raoult's law
`(p^(@)-p_(s))/(p^(@))=x_(B)`
`x_(B)=(1.2-0.6)/(1.2)=0.50`
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