(B) Let $A$ be toluene and $B$ be benzene.Given: $X_{A} = 0.50$,$X_{B} = 1 - 0.50 = 0.50$.$P_{A}^{0} = 37.0 \ torr$,$P_{B}^{0} = 119 \ torr$.Total vap

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(B) Let $A$ be toluene and $B$ be benzene.Given: $X_{A} = 0.50$,$X_{B} = 1 - 0.50 = 0.50$.$P_{A}^{0} = 37.0 \ torr$,$P_{B}^{0} = 119 \ torr$.Total vapour pressure $P_{total} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B} = (37.0 \times 0.50) + (119 \times 0.50) = 18.5 + 59.5 = 78.0 \ torr$.According to Dalton's Law,the mole fraction of toluene in the vapour phase $(y_{A})$ is given by $y_{A} = \frac{P_{A}}{P_{total}}$.$P_{A} = P_{A}^{0} X_{A} = 37.0 \times 0.50 = 18.5 \ torr$.$y_{A} = \frac{18.5}{78.0} \approx 0.237$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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

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A
$0.462$
B
$0.237$
C
$0.506$
D
$0.325$

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

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Two liquids $A$ and $B$ have $P_A^o$ and $P_B^o$ in the ratio of $1 : 3$ and the ratio of number of moles of $A$ and $B$ in the liquid phase is $1 : 3$. The mole fraction of $A$ in the vapour phase in equilibrium with the solution is equal to:

$A$ solution of liquids $A$ and $B$ is given. The mole fraction of $A$ in the vapor phase is $x_1$ and in the liquid phase is $x_2$. If $P_A^o$ and $P_B^o$ are the vapor pressures of pure $A$ and $B$ respectively,then the total vapor pressure $P_{total}$ will be:

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