Question

When 0.4 g of acetic acid is dissolved in 40 g of benzene, the freezing point of the solution is lowered by 0.45 K. Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene.
(K_f for benzene = 5.12 K kg mol⁻¹, at. wt. C = 12, H = 1, O = 16)

Answer 1

∆T_f = 0.45 K 

i = ?

K_f = 5.12 K kg mol⁻¹

M_B of CH₃COOH = 60

W_A = 40 g

W_B = 0.4 g

We know that

`Delta T_f = i K_f * W_B/M_B xx 1000/W_A`

`0.45 = i xx 5.12 xx 0.4/60 xx 1000/40`

i = `(0.45 xx 60 xx 40)/(0.4 xx 1000 xx 5.12)`

= `1080/2048`

= 0.5273

Let degree of association of acetic acid = α

∴ α = `(i - 1)/(1/(n - 1)`  For CH₃COOH, n = 2

= `(0.5273 - 1)/(1/2 - 1)`

= `0.4727/0.5`

= 0.945

∴ Degree of association of acetic acid (α) = 0.945 or 95.5%