A solution at 20°C is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively :

Question

A solution at 20°C is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively : (A) 35.0 torr and 0.480 (B) 38.0 torr and 0.589 (C) 30.5 torr and 0.389 (D) 35.8 torr and 0.280

Tardigrade Admin · Accepted Answer

P text mix = PA ° × A+PB ° XB=74.7 × (1.5/5)+22.3 × (3.5/5) =38 . And y text benzene=(22.41/38)=0.589

35.0 torr and 0.480

38.0 torr and 0.589

30.5 torr and 0.389

35.8 torr and 0.280

Pmix ​=PA∘​×A+PB∘​XB​=74.7×51.5​+22.3×53.5​ =38. And ybenzene​=3822.41​=0.589

$P_{mix} = P_{A}^{\circ} \times A + P_{B}^{\circ} X_{B} = 74.7 \times \frac{1.5}{5} + 22.3 \times \frac{3.5}{5}$

$= 38.$ And $y_{benzene} = \frac{22.41}{38} = 0.589$