At 80,^{circ}C,the vapour pressure of pure li | vedclass

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(D) Given that the solution boils at $80\,^{\circ}C$ and $1\, atm$ pressure,the total vapour pressure $P_{total}$ is equal to the external pressure,wh

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Given that the solution boils at $80\,^{\circ}C$ and $1\, atm$ pressure,the total vapour pressure $P_{total}$ is equal to the external pressure,which is $760\, mm\, Hg$.According to Raoult's law: $P_{total} = X_{A} P_{A}^{\circ} + X_{B} P_{B}^{\circ}$.Since $X_{B} = 1 - X_{A}$,we have $P_{total} = X_{A} P_{A}^{\circ} + (1 - X_{A}) P_{B}^{\circ}$.Substituting the given values: $760 = X_{A} 	imes 520 + (1 - X_{A}) 	imes 1000$.$760 = 520 X_{A} + 1000 - 1000 X_{A}$.$760 = 1000 - 480 X_{A}$.$480 X_{A} = 1000 - 760 = 240$.$X_{A} = \frac{240}{480} = 0.5$.Therefore,the mole percentage of '$A$' is $0.5 	imes 100 = 50\, \%$.

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