At 80,^oC, the vapour pressure of pure liquid | vedclass

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(D) At $1\, atm$ atmospheric pressure,the boiling point of the mixture is $80\,^oC.$At the boiling point,the total vapour pressure of the mixture,$P_{

Vedclass · Accepted Answer

$50$

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(D) At $1\, atm$ atmospheric pressure,the boiling point of the mixture is $80\,^oC.$At the boiling point,the total vapour pressure of the mixture,$P_{T} = 1\, atm = 760\, mm\,Hg.$Using Raoult's law,$P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}.$Given $P_{A}^{\circ} = 520\, mm\,Hg,$ $P_{B}^{\circ} = 1000\, mm\,Hg,$ and $X_{A} + X_{B} = 1,$$760 = 520 X_{A} + 1000(1 - X_{A}).$$760 = 520 X_{A} + 1000 - 1000 X_{A}.$$480 X_{A} = 240.$$X_{A} = \frac{240}{480} = 0.5.$Therefore,the amount of $'A'$ in the mixture is $50\, mol\, percent.$

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

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