Question

At 90°C, the vapour pressure of toluene is 400 torr and that of σ−xylene is 150 torr. What is the composition of the liquid mixture the boils at 90°C, when the pressure is 0.50 atm? What is the composition of vapour produced?

Answer 1

Let, A is Tolune and B is σ−xylene

\(P_A^0 = 400 \ torr. P_B^0 = 150\ torr\)

\(P = 0.50\ atm = 760 \times 0.5 = 380\ torr\)

Using Raoult's law,

\(P = P_A^0 x_A + P_B^0 (1 - x_A)\)

Putting value of P, we get

x_A = 0.92

x_B = 1 − 0.92 = 0.08

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