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

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 `A` and `B` boils at `80@C` and `1 atm` pressure, the amount of `A` in the mixture is `(1 atm =760 mm Hg)`
a. `50 mol %` , b.`52 mol %` ,c.`34 mol%` ,d.`48 mol %`

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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