An aqueous solution of `2` per cent `(wt.//wt)` 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?

Explanation

Vapour pressure of pure water at the boiling point `(p^(@))=" 1 atm = 1.013 bar"`
Vapour pressure of solution `(p_(s))="1.004 bar , Mass of solute "(w_(2))=2 g`
`"Mass of solution = 100 g, Mass of solvent = 98 g"`
Applying Raoult's law for dilute solution `"(being %)"=(p^(@)-p_(s))/(p^(@))-(n_(2))/(n_(1))~=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2))/(M_(2))xx(M_(1))/(w_(1))`
`therefore((1.013-1.004))/("1.013 bar")=(2g)/(M_(2))xx("18 g mol"^(-1))/("98 g")" or "M_(2)=(2xx18)/(98)xx(1.013)/(0.009)" g mol"^(-1)="41.35 g mol"^(-1)`