Question

At ${90}^{o}C$, the vapour pressure of toluene is 400 mm and that of xylene is 150 mm. What is the composition of liquid mixture with respect to xylene (write answer after multiplying with 100) that will boil at ${90}^{o}C$ when the pressure of mixture is 0.5 atm?

Answer 1

The vapor pressure of the mixture at its boiling point $=0.5atm=760/2mm$

$\because P_m=P_T^0\cdot X_T+P_{Xylene}^0\cdot X_{Xylene}$

$\therefore 760/2=400\left(X_T\right)+150(1-X_T)(\because X_T+X_{Xylene}=1)$

$\therefore X_{Toluene}=0.92$ and $X_{Xylene}=1-0.92=0.08$

Hence, the composition (mole fractions) of liquid mixture is 0.92 toluene and 0.08 xylene.