Aqueous solution of 3% non-volatile solute exerts a pressure of 1 bar at the boiling point of the solvent. Calculate the molar mass of solute.

Explanation

Data: (i) Vapour pressure of water at its [boiling point temperature]`=P^(@)=1.013` bar
(ii) Vapour pressure of solution at normal boiling point P=1bar
3% solution=3g of solute in 100g of solution.
(iii) Mass of solute=3g (iv) Mass of solvent=100-3=97g
(v) Relative lowering of vapoure pressure`=(P^(@)-P)/(P)=(1.013-1)/(1.013)=0.0128`
(vi) Molar mass of water=18g `mol^(-1)`
Formula, `M_(2)=(W_(2))/(W_(1))xx(M_(1))/([(P^(@)-P)/(P^(@))])`
Substitution: `M_(2)=(3)/(97)xx(18)/(0.0128)`
answer `M_(2)=43.49g//mol`.