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  • An azeotropic mixture of benzene and toluene boils at and has a c

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 g


Option: 2

68 g
 


Option: 3

70 g

 


Option: 4

72 g


Answers (1)

best_answer

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.

 

Posted by

himanshu.meshram

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