A solution containing 2.5 g of a non-volatile solute in 100 gm of benzene boiled at a temperature 0.42K higher than at the pure solvent boiled. What is the molecular weight of the solute? The molal elevation constant of benzene is `"2.67 K kg mol"^(-1)`.

Explanation

`K_(b)="2.67 K kg mol"^(-1)`
`DeltaT_(b)=0.42K`
`W_(1)=100g=(100)/(1000)kg=0.1Kg`
`W_(2)=(K_(b))/(DeltaT_(b)).(W_(2))/(W_(1))`
`M_(2)=(2.67)/(0.42)xx(2.5)/(0.1)`
`M_(2)="158.98 g mol"^(-1)`