Question

When 2 g of benzoic acid (C₆H₅COOH) is dissolved in 25 g of benzene, it shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.7 K kg mol⁻¹.

What is the percentage association of acid if it forms a dimer in solution?

Answer 1

Given: Weight of solute (w) = 2.0 g

Weight of solvent (W) = 25 g

ΔT_f  = 1.62 K

K_f = 4.7 K kg mol⁻¹

`M_"normal"` = 122 g mol⁻¹

`m_(obs) = (1000 xx K_f xx w)/(ΔT_f xx W)`

= `(1000 xx 4.7 xx 2)/(1.62 xx 25)`

= `(9400)/(40.5)`

= 232.09 g mol⁻¹

= 232.1 g mol⁻¹

i = `("Normal mol. wt.")/("Observed mol. wt.")`

= `122.0/232.1`

= 0.5256 ≈ 0.526

Degree of dissociation (α) =`(1 - i)/(1 - 1/h)`

= `(1 - 0.526)/(1 - 1/2)`

= `0.474/(1/2)`

= 0.474 × 2

= 0.948

= 94.8%