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 = 1
atm = 1.013 bar .
Vapour pressure of solution = 1.004 bar
Mass of solute = 2 g
Mass of solution = 100 g
Mass of solvent = 100 - 2 = 98 g
Applying Raoult.s law
`(p^(@) - p_(s))/(p^(@)) = x_(2) ~~ (W_(2) xxM_(1))/(M_(2)xxM_(1))`
`(1.013 - 1.004)/(1.013) = (2xx18)/(M_(2) xx 98)`
or `" "M_(2) = (2xx 18 xx 1.013)/(0.009 xx 98) = 41.35 g mol^(-1)`