An aqueous solution of `2` % `(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 poin `P^(@))=1.013` bar
Vapour pressure of solution `(P_(s))=1.004` bar
Mass of solute `(w_(2))=2g`
Molar mass of solvent, water `(M_(1))=18 g`
Mass of solvent `(w_(1))=98 g`
Mass of solution=100 g
Applying Raoult's Law for dilute solutions,
`(^(@)-P_(s))/(P^(@))=(n_(2))/(n_(1)+n_(2))=(n_(2))/(n_(1))`
[Dilute solution being 2%]
`(P^(@))-P_(s))/(P^(@))=(n_(2))/(n_(1))=(W_(2)//M_(2))/(W_(1)//M_(1))`
`((1.013)-(1.004))/((1.013))=(2xx18)/(M_(2)xx98)`
` therefore M_(2)=(2xx18)/(98xx0.009)xx1.013=41.35 " g mol"^(-1)`.