\(2.0\, \ g\) of benzoic acid dissolved in \(25\, \ g\) of benzene shows a depression in freezing point equal to \(1.62\, \ K\). Molal depression constant, \(K_f\) of benzene is \(4.9\, \ Kkg\, \ mol^{-1}\). What is the percentage association of the acid?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 99.2%
The depression in freezing point gives the experimental molar mass of the solute:
M2 = (Kf × w2 × 1000)/(ΔTf × w1)
Benzoic acid dimerises in benzene through hydrogen bonds, so the experimental molar mass is larger than the formula mass, and the van’t Hoff factor i = normal molar mass / observed molar mass is less than 1.
M2 = (4.9 × 2.0 × 1000)/(1.62 × 25) = 241.98 g mol−1 ≈ 242 g mol−1
Normal molar mass of \( \ce{C6H5COOH} \) = 122 g mol−1
i = 122/242 = 0.504
If x is the degree of association, for \( \ce{2C6H5COOH} \) ⇌ \( \ce{(C6H5COOH)2} \) one mole gives (1 − x) mol of monomer and x/2 mol of dimer:
i = 1 − x + x/2 = 1 − x/2 = 0.504
x = 2(1 − 0.504) = 0.992, i.e. 99.2%
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Abnormal Molar Masses – “Example 1.12 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms dimer in solution?
Now consider the following equilibrium for the acid: 2 C6H5COOH ⇌ (C6H5COOH)2
Therefore, degree of association of benzoic acid in benzene is 99.2 %.”