Question

A solution has 1:4mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20°C are 440Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in vapour phase would be

Answer 1

Let the pentane be 1 moleand hexane be 4 mole.
Total number of moles of pentane and hexane in mixture = 1 + 4 = 5 mol
Mole fraction of pentane in solution = 1/5
Mole fraction of hexane in solution = 4/5
Total pressure of solution :
Ps = XP PPo + XH PHo
= 1/5 (440 mm Hg) + 4/5 (120 mm Hg)
= 88 mm Hg + 96 mm Hg
= 184 mm Hg
Mole fraction of pentane in vapour phase = XP PPo/ Ps ​= 88 / 184= 0.478