The freezing point of a solution containing 0.2g of acetic acid in 20.0g benzene is lowered by 0.45oC. The degree of association of acetic acid in benzene is: [Assume acetic acid dimerizes in benzene and Kf for benzene = 5.12 K kg mol⁻¹]

The molar mass of acetic acid is 60 g/mol.
The molality of the solution is the number of moles of solute divided by the mass of solvent (in kg).
Number of moles of acetic acid = (0.2 g) / (60 g/mol) = 0.00333 mol
Mass of benzene = 20.0 g = 0.02 kg
Molality (m) = (0.00333 mol) / (0.02 kg) = 0.1665 mol/kg
ΔTf = Kf * m * i
where:
ΔTf = freezing point depression = 0.45 oC
Kf = cryoscopic constant for benzene = 5.12 K kg mol⁻¹
m = molality = 0.1665 mol/kg
i = van't Hoff factor (accounts for the dissociation or association of solute)
0.45 oC = 5.12 K kg mol⁻¹ * 0.1665 mol/kg * i
i = 0.45 / (5.12 * 0.1665) = 0.526
If acetic acid did not dimerize, i would be 1. Since it dimerizes, the i value will be less than 1.
Let α be the degree of association. The dimerization reaction is:
2CH3COOH <=> (CH3COOH)2
Initial moles: 1 0
At equilibrium: 1-α α/2
Total moles at equilibrium = 1 - α + α/2 = 1 - α/2
i = (1 - α/2) / 1 = 1 - α/2
0.526 = 1 - α/2
α/2 = 1 - 0.526 = 0.474
α = 0.948 ≈ 94.8%
Therefore, the degree of association of acetic acid in benzene is approximately 94.5%.