The molar mass of benzoic acid in benzene, as determined by the depression in freezing point method, corresponds to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Dimerisation of benzoic acid
Depression in freezing point depends on the number of solute particles. In benzene, a solvent of low dielectric constant, two benzoic acid molecules join through hydrogen bonds between their –COOH groups:
\( \displaystyle \ce{2C6H5COOH} \rightleftharpoons \ce{(C6H5COOH)2} \)
The number of particles is nearly halved (i ≈ 0.5), so ΔTf is about half the expected value and the molar mass worked out from it is about twice the true value of 122 g mol−1. The molar mass found therefore corresponds to the dimer.
Why the other options are incorrect
Option (1) Ionisation of benzoic acid — Incorrect. Ionisation would increase the number of particles and give a molar mass lower than the true value, not a higher one.
Option (3) Trimerisation of benzoic acid — Incorrect. Hydrogen bonding joins the molecules in pairs; a trimer would give about three times the true molar mass.
Option (4) Solvation of benzoic acid — Incorrect. Solvation does not change the number of solute particles, so it cannot make the molar mass come out abnormal.
NCERT (2026-27 edition):
1.7 Abnormal Molar Masses – “… Molecules of ethanoic acid (acetic acid) dimerise in benzene due to hydrogen bonding. This normally happens in solvents of low dielectric constant. In this case the number of particles is reduced due to dimerisation. … It can be undoubtedly stated here that if all the molecules of ethanoic acid associate in benzene, then ΔTb or ΔTf for ethanoic acid will be half of the normal value. The molar mass calculated on the basis of this ΔTb or ΔTf will, therefore, be twice the expected value.”
Example 1.12 – “… Now consider the following equilibrium for the acid:
\( \displaystyle \ce{2C6H5COOH} \rightleftharpoons \ce{(C6H5COOH)2} \)
… Therefore, degree of association of benzoic acid in benzene is 99.2 %.”