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The vapour pressure of pure liquid solve...

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

A

`0.50`

B

`0.25`

C

`0.75`

D

gives data is not sufficient

Text Solution

Verified by Experts

The correct Answer is:
B
`(P^(@) -P_(s))/(P^(@)) = X_("solute")`
`X_("solute") = 0.25`
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