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

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(B) The boiling point of a mixture is the temperature at which its total vapour pressure equals the external pressure.Given: $P_{A}^{o} = 520\,mmHg$,$

Vedclass · Accepted Answer

$50$

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(B) The boiling point of a mixture is the temperature at which its total vapour pressure equals the external pressure.Given: $P_{A}^{o} = 520\,mmHg$,$P_{B}^{o} = 1000\,mmHg$,$P_{T} = 1\,atm = 760\,mmHg$.According to Raoult's Law: $P_{T} = P_{A}^{o} \chi_{A} + P_{B}^{o} \chi_{B}$.Since $\chi_{A} + \chi_{B} = 1$,we have $\chi_{B} = 1 - \chi_{A}$.Substituting the values: $760 = 520 \chi_{A} + 1000(1 - \chi_{A})$.$760 = 520 \chi_{A} + 1000 - 1000 \chi_{A}$.$480 \chi_{A} = 240$.$\chi_{A} = \frac{240}{480} = 0.5$.Therefore,the mole percentage of $A$ is $0.5 	imes 100 = 50\,mol\,\%$.

At $80\,^oC$,the vapour pressure of pure liquid $A$ is $520\,mmHg$ and that of pure liquid $B$ is $1000\,mmHg$. If a mixture of $A$ and $B$ boils at $80\,^oC$ and $1\,atm$ pressure,the amount of $A$ in the mixture is $........$ $mol\,\%$.

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