`1.22 g` of benzoic acid is dissolved in acetone and benzene separately. Boiling point of mixture with acetone increase by `0.17^(@)C` and boiling point of mixture with benzene increases by `0.13^(@)C`.
`K_(b) ("acetone")=1.7 K kg "mol"^(-1)`,
Mass of acetone `= 100 g`,
`K_(b) ("benzene")=2.6 k Kg mol^(-1)`,
Mass of benzene `= 100 g`,
Find molecular weight of benzoic acid in acetone and in benzene solution. Justify your answer with structure.

Answer

`122,224`

Explanation

(i) In first case,
`DeltaT_(b)=K_(b)xxm=K_(b)xx("Wt. of solute")/("Mol.wt. of solute")xx(1000)/(" wt. of solvent")`
or `0.17=1.7xx(1.22)/(Mxx100xx10^(-3))`
`M=122gm//"mole"`
Thus the benzoic acid exists as a monomer in acetone
(ii) In second case,
`DeltaT_(b)=K_(b)xx("Wt. of solute")/("Mol. wt. of solute")xx(1000)/("wt. of solvent")`
or `0.13=2.6xx(1.22)./(Mxx100xx10^(-3))`
`M=224`
Double molecular weight of benzoic acid `(244)` in acetone solution indicates that benzoic acid exists as dimer in acetone.