Question

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

Answer 1

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.