An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?
[Vapour pressure of water = 1.013 bar] 

41.3 g mol^-1

An aqueous solution of 2 percent non-volatile solute means 2 g of solute is present in 100 g of solution.

Weight of solvent = 100 - 2 = 98 g

\(\frac{P^o - P}{P^o}\) = \(\frac{W_B}{M_B}\) x \(\frac{M_A}{W_A}\)

\(\frac{1.013 - 1.004}{1.013}\) = \(\frac{2}{M_B}\) x \(\frac{18}{98}\)

M_B = 41.3 g mol^-1