An aqueous solution of 2 % non - volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solven. What is the molar mass of the solute ?

Explanation

Here,
Vapour pressure of the solution at normal boiling point `(p_(1))=1.004` bar
Vapour pressure of pure water at normal boiling point, `(p_(1)^(0))=1.013` bar
Mass of solute, `(w_(2))=2 g`
Mass of solvent (water), `(w_(1))=98 g`
Molar mass of solvent (water), `(M_(1))=18 g mol^(-1)`
According to Raoult.s law,
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=(w_(2)xxM_(1))/(M_(2)xx w_(1))`
`(1.013-1.004)/(1.013)=(2xx18)/(M_(2)xx98)`
`M_(2)=(1.013xx2xx18)/(0.009xx98)`
`= 41.35 g mol^(-1)`
Hence, the molar mass of the solute is 41.35 g `mol^(-1)`.