a 2g of benzoic acid dissolve in 25g of benzene shows a depression in freezing point of 1.62 k.

molal depression constant for benzene is 4.9 KKg/mol. what is the % of association of acid.

if it forms a dimer in solⁿ .

Given: 

W_A = 2 g

W_B= 25 g,

K_f = 4.9 K kg mol^–1

T_f = 1.62 K

Molar mass of the solute can be found out using the equation:

=

M_B = 4.9 K kg mol^–1 × 2g × 1000 g kg ^–1/1.62 K × 2g = 241.98 g mol^–1

Therefore, molar mass of benzoic acid in benzene is= 241.98 g mol^–1

Considering the following equilibrium for the acid:

2C₆H₅COOH ⇔ (C₆H₅COOH)₂

If x represents the degree of association of the solute then we would have (1 – x ) mol of benzoic acid left in unassociated form and x/2 as associated moles of benzoic acid at equilibrium. Correspondingly x/2 as associated moles of benzoic acid at equilibrium.

Therefore, total number of moles at equilibrium will be:

= 1−x+x/2 

1-x/2 = i

Total number of moles of particles at equilibrium equals van’t Hoff factor i.

We know that,

i = Normal molar mass / Abnormal molar mass

= 122 g mol^-1/241.98 g mol

1-x/2 = i = 122 g mol^-1/241.98 g mol

or x/2 = 1 – 122 /241.98 = 1 – 0.0504 = 0.496

or x = 2 × 0 .496 = 0.992

Therefore, degree of association of benzoic acid in benzene is 99.2 %.