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At 80^(@)C, the vapour pressure of pure ...

At `80^(@)C`, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture of solution of 'A' and 'B' boils at `80^(@)C` and 1 atm pressure, the amount of 'A' in the mixture is ( 1 atm `=760` mm Hg.)

Text Solution

Verified by Experts

`P_(M)=p_(A)^@chi_(A) + p_(B)^@chi_(B)`
`P_(M)=p_(A)^@chi_(A) + p_(B)^@(1-chi_(A))`
`760=520chi_(A) + 1000-1000chi_(A)`
`chi_(A)=240/480 = 0.5`
Therefore, `mol% = 50`
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