InterviewSolution
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InterviewSolution
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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 `[email protected]` 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 %`A. `52` mol per centB. `34` mol per centC. `48` mol per centD. `50` mol per cent |
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Answer» Correct Answer - D `P_(m) = P_(A)^(@)X_(A) + P_(B)^(@)X_(B)` `760 = 520 X_(A) + P_(B)^(@)(1-X_(A))` `:. X_(A) = 0.5 = 50%` |
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