Question

A solution is prepared by dissolving 32.5 g of a nonvolatile solute in 200 g of water. The vapor pressure above the solution is 21.85 torr and the vapor pressure of pure water is 23.76 torr at this temperature. What is the molecular weight of the solute?

Answer 1

The molecular weight of the solute will be 33.6 grams

Explanation :

According to Raoult's Law,

Pwater = Xwater x P°water

=> X = P/P°

=> X = 21.85/23.76

=> X = 0.92

mass of water in the solution = 200 g

=> moles of water in the solution = 200/18 = 11.11

let the number of moles of solute = n

=> mole fraction of water = 0.92

=> 11.11/(11.11 + n) = 0.92

=> 11.11 = 10.22 + 0.92n

=> 0.92n = 0.89

=> n = 0.89/0.92 = 0.967

=> mass/molecular mass = 0.967

=> 32.5/M = 0.967

=> M = 33.6  grams

hence the molecular weight of the solute will be 33.6 grams