Vapour pressure of pure benzene is 119 torr | vedclass

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(B) Let $A$ be benzene and $B$ be toluene. Given: $P_A^o = 119 \ torr$,$P_B^o = 37.0 \ torr$,$x_B = 0.50$. Since $x_A + x_B = 1$,$x_A = 0.50$. Accordi

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$0.237$

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(B) Let $A$ be benzene and $B$ be toluene. Given: $P_A^o = 119 \ torr$,$P_B^o = 37.0 \ torr$,$x_B = 0.50$. Since $x_A + x_B = 1$,$x_A = 0.50$. According to Raoult's Law,the partial pressures are: $P_A = P_A^o x_A = 119 	imes 0.50 = 59.5 \ torr$ and $P_B = P_B^o x_B = 37.0 	imes 0.50 = 18.5 \ torr$. The total pressure $P_{total} = P_A + P_B = 59.5 + 18.5 = 78.0 \ torr$. In the vapour phase,the mole fraction of toluene $(y_B)$ is given by Dalton's Law: $y_B = \frac{P_B}{P_{total}} = \frac{18.5}{78.0} \approx 0.237$.

Vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature. The mole fraction of toluene in the vapour phase,which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$,will be:

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