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0.900g of a solute was dissolved in 100 ...

0.900g of a solute was dissolved in 100 ml of benzene at `25^(@)C` when its density is 0.879 g/ml. This solution boiled `0.250^(@)C` higher than the boiling point of benzene. Molal elevation constant for benzene is `"2.52 K.Kg.mol"^(-1)`. Calculate the molecular weight of the solute.

Text Solution

Verified by Experts

`"Weight of benzene "=100xx0.879=87.9g`
Molality of solution, m `=(0.900//M_(2))/(87.9)xx1000`
`=(900)/(87.9M_(2))`
`DeltaT_(b)=K_(b)m" (or) 0.52"xx(900)/(87.9M_(2))`
`therefore" "M_(2)=(900xx2.52)/(87.9xx0.250)`
`M_(2)="103.2g/mole"`
`therefore" "{:("Molecular weight"),("of the solute"):}}="103.2 g/mole"`
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