Question

2 g of benzoic acid dissolved in 25 g of C₆H₆ show a depression in freezing point equal to 1.62 K. Molal depression constant of C₆H₆ is 4.9 K mo1⁻¹ kg. What is the percentage association of acid if it forms double molecules in solution?

Answer 1

In the present case, w = 2 g, W = 25 g, ΔT_f = 1.62 K, K_f = 4.9 K kg mol⁻¹

The observed molecular mass of benzoic acid can be calculated from the formula 

`M'_"obs" = (1000 xx K_f xx w)/(W xx Delta T_f)`

= `(1000 xx 4.9 xx 2)/(25 xx 1.62)`

= 241.975

Normal molecular mass of benzoic acid (C₆H₅COOH) = 122

∴ `i = "Normal molecular mass"/"Observed molecular mass"`

= `122/241.975`

= 0.504    ...(i)

Further in solution, if a is the degree of association of benzoic acid, we have

	\[\ce{2C6H5COOH <=> (C6H5COOH)2}\]
Initially	1 mol                                        -
At equilibrium	(1 − α)                                     `alpha/2`

∴ Total number of moles in solution = `1 - alpha + alpha/2`

= `1 - alpha/2`

`i = "No. of moles in solution"/"No. of moles added"`

= `(1 - alpha/2)/1`

= `1 - alpha/2`    ...(ii)

Comparing eqs. (i) and (ii), we have

`1 - alpha/2 = 0.504`

∴ α = (1 − 0.504) × 2

= 0.992

Hence, the per cent association of benzoic acid in benzene = 0.992 × 100 = 99.2%