Question

At ${90}^{\circ }C,$ the vapour pressure of toluene is 400 torr and that of $\sigma -xylene$ is 150 torr. What is the composition of the liquid mixture the boils at ${90}^{\circ }C,$ when the pressure is 0.50 atm? What is the composition of vapour produced?

• A. Toluene$8mol$%, Toluene=$3.2mol$%
• B. Toluene$15mol$%, Toluene=$36mol$%
• C. Toluene$85mol$%, Toluene=$64mol$%
• D. Toluene$92mol$%, Toluene=$96.8mol$%

Answer 1

Answer: (D)

Toluene$92mol$%, Toluene=$96.8mol$%

Let, A is $Tolune$ and B is $\sigma -xylene$

$P_A^o=400torr$, $P_B^o=150torr$

$P=0.50atm=760\times 0.5=380torr$

Using Raoult's law,

$P=P_A^o{x}_A+P_B^o(1-x_A)$

Putting value of P from equation (1) we get

$x_A=0.92$

$x_B=1-0.92=0.08$

Hence, $x_A=92$ mol percent $x_B=8$ mol percent

So, option D is correct