The total vapour pressure of a solution of li | vedclass

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(C) Given: Total pressure $P_S = 600 \ torr$,mole fraction of $A$ in liquid phase $X_A = 0.7$,mole fraction of $B$ in liquid phase $X_B = 1 - 0.7 = 0.

Vedclass · Accepted Answer

$300 \ torr, 1300 \ torr$

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(C) Given: Total pressure $P_S = 600 \ torr$,mole fraction of $A$ in liquid phase $X_A = 0.7$,mole fraction of $B$ in liquid phase $X_B = 1 - 0.7 = 0.3$,and mole fraction of $A$ in vapour phase $Y_A = 0.35$. Using Dalton's Law,the partial pressure of $A$ is $P_A = Y_A 	imes P_S = 0.35 	imes 600 = 210 \ torr$. From Raoult's Law,$P_A = P_A^o 	imes X_A$,so $P_A^o = \frac{210}{0.7} = 300 \ torr$. Since $P_S = P_A + P_B$,we have $P_B = 600 - 210 = 390 \ torr$. Using Raoult's Law for $B$,$P_B = P_B^o 	imes X_B$,so $P_B^o = \frac{390}{0.3} = 1300 \ torr$. Therefore,the vapour pressures of pure $A$ and $B$ are $300 \ torr$ and $1300 \ torr$ respectively.

The total vapour pressure of a solution of liquid $A$ and liquid $B$ is $600 \ torr$. The mole fraction of component $A$ in the liquid phase is $0.7$. If the mole fraction of $A$ in the vapour phase is $0.35$,what are the vapour pressures of pure $A$ and pure $B$ respectively?

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