An aquase solution of 2% non volatile solute exerts a pressure of 1.004 bar at normal boiling point of solvent. What is the molecular mass of solute (vapour pressure of pure water is 1.013bar

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar (Given)

Vapour pressure of pure water at normal boiling point (p10) = 1.013 bar

Mass of solute, (w2) = 2 g
Mass of solvent (water), (w1) = 100 - 2 = 98 g
Molar mass of solvent (water), (M1) = 18 g mol - 1
According to Raoult's law,
(p10 - p1) / p10 = (w2 x M1 ) / (M2 x w1 )

(1.013 - 1.004) / 1.013 = (2 x 18) / (M2 x 98 )

0.009 / 1.013 = (2 x 18) / (M2 x 98 )
M2 = (2 x 18 x 1.013) / (0.009 x 98)
M2 = 41.35 g mol - 1
Hence, the molar mass of the solute is 41.35 g mol - 1.