The freezing point of benzene decreases by 0.45oC when 0.2g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be :(Kf for benzene=5.12 K kg mol⁻¹)

Molality m = (1000g/kg × 0.2g)/(60g/mol × 20g) = 0.1667 m
ΔTf = iKfm
0.45 = i × 5.12 × 0.1667
The van't Hoff factor i = 0.527
The degree of association α = (1 - i)/(1 - 1/n)
The degree of association α = (1 - 0.527)/(1 - 1/2) = 0.946
Percent association = 0.946 × 100 = 94.6
Hence, the option (C) is the correct answer.