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Q)

Mole fraction of the toluene in the vapour phase which is in equilibrium with a solution of benzene $(p^0=120torr)$ and toluene$(p^0=80torr)$ having 2.0mole of each is

$\begin{array}{1 1}(a)\;0.40\\(b)\;0.60\\(c)\;0.25\\(d)\;0.50\end{array}$

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A)
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• 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 $p_T = p_1 + p_2 = x_1p_1^0 + x_2p_2^0$
Partial pressure of Benzene= $p_1 =x_1 p_1^0 = \frac{2}{2+2} \times 120\; torr = 0.5 \times 120\; torr$
Partial pressure of Toulene= $p_2 =x_1 p_1^0 = \frac{2}{2+2} \times 80\; torr = 0.5 \times 80\; torr$
According to Dalton's law of partial pressures, total pressure over the solution $p_T = p_1 + p_2 = x_1p_1^0 + x_2p_2^0$
$\therefore$ Total pressure $p = 0.5 \times (120+80) = 100\; torr$
At equilibrium, if $x$ be the mole fraction of toulene in vapour phase,
Partial pressure of Toulene = Total pressure of the solution $\times$ mole fraction of toulene in vapour pressure
$\Rightarrow 80\; torr \times 0.5 = p \times x$
$\Rightarrow 80\; torr \times 0.5 = 100\; torr \times x$
$\Rightarrow x = 0.4$