A certain solution of benzoic acid in benzene boil at `82.6^(@)C` and freezes at `3.1^(@)C`. What information about the no of particiles and the structure of benzoic acid at the two temperature can be deduced from the above data ? The boiling point and freezing point of pure benzene are `180.1^(@)C and 5.5^(@)C` respectively.
`K_(r)=5.12 k kg//"mol", k_(b)=2.67 k` kg/ mol.

Explanation

Depression in freezing point
`=5.53-1=2.4`
Elveation in boiling point `=82.6-80.1=2.5`
We know that
`DeltaT_(f)=k_(f)xxm`
`rArr m_(1)=(Delta_(f))/(K_(f))=(2.4)/(5.12)=0.468`
And similarly
`DeltaT_(b)=K_(b)xxm_(2)`
`rArr m_(r)=(2.5)/(2.67)=0.936`
From the above calculation it is clear that the molality is half at freezing point as compared to molality at boiling point or we can say that the no. of mols of benzoic acid at its freeing point is half of that at boiling pont, this is possible only when benzoic acid dimerises at low temperature.