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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

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The correct Answer is:
0.25
`X_(B)=(P_(A)^(0)-P_("sol"))/P_(A)^(0)=(0.80-0.60)/0.80=0.25`
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