Question

What is the freezing point of 0.4 molal solution of acetic acid in benzene in which it dimerises to the extent of 85%. Freezing point of benzene is 278.4 k and its molar heat of fusion is 10.042 kjmol-1.

Answer 1

Given :

1) Molality of acetic acid in benzene, m = 0.4

2) Extent of dimerization, α = 85%

3) Freezing point of benzene Tf = 278.4 K

4) Molar heat of fusion = 10.042 kJ/mol

Molar mass of acetic acid, M = 78

To find :

Freezing point of acetic acid

Solution :

• We know that, Kf i.e molal depression constant is given as below

$Kf = R \times Tf^{2} \times M / 1000 \times \: Hf$

• Substituting values of R, Tf, M and Hf in above formula

Kf = 8.314×(278.4)^2×78 / 1000×10.042×10^3

Kf = 5 K Kg/mol

• Also given that acetic acid dimerises to 85% extent

Therefore,

α = i - 1 / (1/n) - 1

= i - 1 / (1/2) - 1

0.85 = i - 1 / (-1/2)

i - 1 = - 0.85×1 / 2 = - 0.425

i = 0.575

We also know that, ∆Tf is given as

∆Tf = i×m×Kf

∆Tf = 0.575×0.4×5

∆Tf = 1.15

Tf' = 278.4 - 1.15

Tf' = 277.25 K

Answer 2

Hence the freezing point of the benzene is Tf = 277.2 K

Explanation:

K f = RT^2 M  /   1000 × ΔH  fusion

Here, K  f  =  8.314 × (278.4)^2  × 78  /  1000 × 10.042 × 10^ 3  

Kf ​= 5 KKg/mol

2CH3  COOH ⟶ (CH 3 COOH)2

α =   i − 1  /  1 / n  − 1 =  i−1  /  1/2 − 1  

α​ = 0.8

​i−1=−0.4

⇒i = 0.6

ΔT  f  = K  f  × m × i

5×0.6×0.4

=1.2

∴T  f 1  = 278.4 − 1.2

Tf = 277.2 K

Hence the freezing point of the benzene is Tf = 277.2 K