The vapour pressure of a solvent decreased by 20 mm of Hg when a non-volatile solute was added to the solvent. The mole-fraction of the solute in the solution is 0 . 5 . What should be the mole-fraction of the solvent for the decrease in the vapour pressure needs to be 10 mm of Hg ?

Step-by-step solution

The vapour pressure of the solute =   P o - P s = = 20   mm   of   Hg Now, P A = P A 0 X A Partial pressure of solute P A = P A 0 X A = P A 0 × mole   fraction   of   solute P A 0 = 20 0 . 5 = 40 Pressure of pure solute is equal to the 40   mm   of   Hg If it is decreases by 10   mm   of   Hg . P A = P A 0 X A 10 = 40 × X A X A = 0 . 25 So, mole fraction of solvent = 1 - 0 . 25 = 0 . 75 = 3 4 .