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At room temperature the mole fraction of...

At room temperature the mole fraction of a solution is 0.25 and the vapour pressure of the solvent is 0.80 atm. Then the lowering of vapour pressure is

A

0.75

B

0.6

C

0.2

D

0.8

Text Solution

Verified by Experts

The correct Answer is:
C
`(P^(@)-P_(s))/(P^(@)) = x_(2) :. (P^(@)-P_(s))/(0.80) = 0.25`
or `P^(@)-P_(s) = 0.20` atm
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