A certain solution of `1 m` benzoic acid in benzene has a freezing point of `3.1^(@)C` and a normal boiling point of `82.6 ^(@)C`. The freezing point of benzene is `5.5^(@)C`. And its boiling point is `80.1^(@)C`. Analyze the state of the solute (benzoic acid ) at two temperature and comment .

Answer

(a) benzoic acid may ionise
so that `i=(1+x)gt1`
`(##RES_PHY_CHM_V01_XII_C02_E01_314_A01##)`
`(##RES_PHY_CHM_V01_XII_C02_E01_314_A01##)`
`i=[1+((1)/(n)-1)xx]lt1`
(c) benzoic acid remains unchanged so that `i=1`
First we consider depression in freezing point of solution of molality `m`
`DeltaT_(f)=K_(f)mi`
`i=(DeltaT_(f))/(K_(f)m)=(5.5-3.1)/(5.12)=0.47`
We find `ilt1` (case b). Hence, there is association of benzoic acid in benzenen due to intermolecular `H-`bonding in freezing state.
`(##RES_PHY_CHM_V01_XII_C02_E01_314_A01##)`
For elevation in boiling point for the same solution of molality `m,DeltaT_(b)=K_(b)mi`
`i=(DeltaT_(b))/(K_(b)m)=(82.6-80.1)/(2.53)=0.99approx1.0`
we find `i=1` (case c). Hence, benzoic acid in benzene acid in benzenen in boiling state remains unchanged.