3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated). [Given : Molar mass of benzoic acid `= 122 g "mol"^(-1), K_f` for benzene = `4.9 K kg "mol"^(-1) `]

Explanation

Apply the relation, `M_2 = (1000 xx w_2 xx K_f)/(DeltaT_f xx w_1)`
Substituting the values, we have
`M_2 = (1000 g kg^(-1) xx 3.9 g xx 4.9 K kg "mol"^(-1))/(1.62 k xx 49 g) = 240 g "mol"^(-1)`
van.t Hoff factor = Normal molar mass / Abnormal molar mass
` = 122/240 = 0.508`
Benzoic acid gets associated (dimerised).