The value of van't Hoff factor, \( i \), for \( CH_3COOH \) solution in water will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Between 1 and 2
The van’t Hoff factor i is the number of particles a solute actually gives in solution per formula unit, compared with what it would give if it neither dissociated nor associated. i = 1 for a solute that stays as single molecules, i > 1 for dissociation, and i < 1 for association.
Acetic acid is a weak electrolyte. In water only a small fraction of its molecules ionise:
\( \displaystyle \ce{CH3COOH <=> CH3COO- + H+} \)
If the degree of dissociation is α, each mole gives \( (1-\alpha) + \alpha + \alpha = 1 + \alpha \) moles of particles, so i = 1 + α. Since 0 < α < 1, the value of i lies between 1 and 2.
Why the other options are incorrect
Option (2) Less than 1 — Incorrect. This holds for acetic acid in benzene, where it forms hydrogen-bonded dimers (association); in water it dissociates.
Option (3) 2 — Incorrect. i = 2 would mean complete dissociation into two ions, as for a strong electrolyte like NaCl; acetic acid ionises only partly.
Option (4) 1 — Incorrect. i = 1 applies to a non-electrolyte such as glucose or urea, which does not ionise at all.
NCERT (2026-27 edition):
Abnormal Molar Masses – …
