A mole ratio of pentane and hexane in a solution in1:4 the vapour pressure of pure pentane and hexane are 400 and 120nm of Hg calculate the mole fraction of pentane in vapour phase?

Let the pentane be 1 mole and 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 (400 mm Hg) + 4/5 (120 mm Hg)
   = 80 mm Hg + 96 mm Hg
  = 176 mm Hg  

Mole fraction of pentane in vapour phase = XP PPo/ Ps ​
                                                                 = 80 / 176
                                                                 = 0.45454

Regards

Arun (askIITians forum expert)