An aqueous solution of `2%` nonvolatile solute exerts a pressure of `1. 004` bar at the boiling point of the solvent. What is the molar mass of the solute when `P_(A) ^(@)` is `1.013` bar ?

Explanation

Vapour pressure of pure water at the boiling point.
`(P^(0)) =1 atm = 1.013` bar
Vapour pressure of solution `P _(S) = 1. 004` bar
`M_(1) = 18 g mol ^(-1)`
`M _(2) ` = ?
Mass of solute `= W_(2) = 2g`
Mass of solution = 100 g
Mass of solvent `W_(1) = 98 g`
Applying Ranoult.s law for dilute solution
`(P ^(0) -P_(S))/(P ^(0)) = (n_(2))/(n _(1) + n _(2)) = (n_(2))/( n _(1))= ((W _(2))/( M _(2)))/( (W _(1))/( M _(2)))`
`(P ^(0) -P_(S))/(P ^(0))= (W_(2))/(M _(2)) xx (M_(1))/(W _(1))`
`(1.013-1,004)/(1.013) = (W_(2))/(M _(2)) xx (18 mol ^(-1))/(98g) (or)`
`M _(2) = (2 xx 18 xx 1. 103)/(98 xx 0.009) = 41.35g mol ^(-1)`