A solution at 20 °C is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressures of pure benzene and toluene are 74.7 Torr and 22.3 Torr, respectively, then the total vapour pressure of the solution and benzene mole fraction in equilibrium with it will be respectively,

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35.8 Torr and 0.280

38.0 Torr and 0.589

30.5 Torr and 0.389

35.0 Torr and 0.480

answer is B.

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Detailed Solution

Mole fraction of benzene, $x_{b} = \frac{n_{b}}{n_{b} + n_{t}} = \frac{1.5}{1.5 + 3.5} = 0.3$

Mole fraction of toluene,

$x_{t} = 1 - x_{b} = 0.7$

The vapour pressure of the solution is

$\begin{array}{l} p = x_{b} p_{b}^{*} + x_{t} p_{t}^{*} = (0.3 \times 74.7 + 0.7 \times 22.3) Torr \\ = 38.02 Torr \end{array}$

Mole fraction of benzene in vapour phase is

$y_{b} = \frac{p_{b}}{p} = \frac{(0.3 \times 4.7) Torr}{(38.02 Torr)} = 0.589$

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