Question

At 90°C, the vapor pressure of pure methylbenzene is 400 Torr and that of pure 1,2-dimethylbenzene is 150 Torr. What the composition of a liquid mixture that boils at 90°C when the pressure is 0.50 atm? What is the composition of the vapor produced?

Answer 1

The question is based on Racult's law. methy benzene = 400 tors Dimethyl Benzone = 150 toor p°: vapour poressure of chemical in This composition of liquid mixture that boils at 90c 1) Methylbenzene → 92% 0.92 2) Dimethylbenzare 8.6. 0.08 ture state Ym = PM = PM The mixture boils at 0.50 atm. e total = 0. 50 atm = 380 tors Butk To Tot You: composition in Vapour phase Ym = bom em = 400 X 0.92 = 1 auton 0.5 atm 760 tors 380 tors brot 380 1Ym = 0.9684 Using Raoult's law; Protat Pomer. + Prime Where mi Methyl benzene Di dimethyl benzene. þ Xo = 150 x 0.08 brot 380 Yo = 0.03151 Thes, composition of vapour produced, 1) Methyl benzene 96.84.1 0.9684 2) Dimethylbenzene 3.151 0.0315 kan 400 lm + 150 (1-xm) 1 Total torr Xm Mole fraction of Methyl benzene. Ro: " " " Dimethylbenzene Also, Hm+X0=1 (:: Only 2 components are present) 380 = 1ook toor torr +15o -150xm tors torr Xm = 0.92 Xo=0:08]