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The vapour pressure of a pure liquid is ...

The vapour pressure of a pure liquid is 0.80 atm. When a non-volatile solute is added to this liquid, its vapour pressure drops to 0.60 atm. The mole fraction of the solute in the solution is

A

`0.150`

B

`0.25`

C

`0.50`

D

`0.75`

Text Solution

Verified by Experts

The correct Answer is:
B
`(P^(0)-P)/(P^(0))= X_(B)`
`(0.2)/( 0.8) = 0.25 =X_(B)`
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