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# Mole fraction of the benzene in the vapour phase which is in equilibrium with a solution of benzene $(p^0=120torr)$ and toluene$(p^0=80torr)$ having 4.0mole of benzene and 6 mole of toluene in solution.

$\begin{array}{1 1}(a)\;0.50\\(b)\;0.40\\(c)\;0.30\\(d)\;0.20\end{array}$

Toolbox:
• According to Dalton’s law of partial pressures, the total pressure over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as $p_T = p_1 + p_2 = x_1p_1^0 + x_2p_2^0$
Given, Moles of Benzene = 4.0 moles
Moles of Toulene = 6.0 moles
Partial pressure of Benzene= $p_1 =x_1 p_1^0 = \frac{4}{4+6} \times 120\; torr = 0.4 \times 120\; torr$
Partial pressure of Toulene= $p_2 =x_1 p_1^0 = \frac{6}{4+6} \times 80\; torr = 0.6 \times 80\; torr$
According to Dalton’s law of partial pressures, the total pressure over the solution phase $p_T = p_1 + p_2 = x_1p_1^0 + x_2p_2^0$
Total pressure $p = 120 \times 0.4 + 80 \times 0.6 = 96\; torr$
At equilibrium, if $x$ be the mole fraction of Benzene in vapour phase,
Partial pressure of Benzene = Total pressure of the solution $\times$ mole fraction of Benzene in vapour pressure
Therefore, $80\; torr \times 0.6 = 96 \; torr \times x \Rightarrow x = 0.5$
edited Jul 20, 2014