#### An azeotropic mixture of benzene and toluene boils at $80.1^{\circ} \mathrm{C}$ and has a composition of $65.4 \%$ benzene and $34.6 \%$ toluene. If a solution of  $50 \mathrm{~g}$ benzene and $50 \mathrm{~g}$ toluene is distilled, how much of the azeotropic mixture will be obtained?Option: 1 64 gOption: 2 68 g  Option: 3 70 g  Option: 4 72 g

The given solution contains 50 g of benzene and 50 g of toluene.

To obtain the azeotropic mixture, the solution is distilled until the temperature reaches $80.1^{\circ} \mathrm{C}$ . At this temperature, the vapour produced will have the same composition as the azeotropic mixture, which is $65.4%$  benzene $34.6%$ and toluene.

Let x be the amount of the azeotropic mixture obtained.

Since the composition of the azeotropic mixture is $65.4%$ benzene and $34.6%$ toluene, we can write:

\begin{aligned} & (0.654 x) g \text { benzene }=(50-x) g \text { benzene } \\ & (0.346 x) g \text { toluene }=(50-x) g \text { toluene } \end{aligned}

Solving these two equations, we get:

$x=70.2 g$

Therefore, approximately 70 g of the azeotropic mixture will be obtained. Hence, the correct answer is option C.