Arrange the following equimolar solutions in the increasing order of…

The correct answer is Option (3) → D, A, C, B

The van’t Hoff factor (i) compares the number of particles actually present in solution with the number of formula units dissolved:

    \( \displaystyle i=\frac{\text{number of particles after dissociation or association}}{\text{number of formula units dissolved}} \)

• Dissociation (electrolytes in water) gives more particles, so i > 1.

• Association (molecules pairing up) gives fewer particles, so i < 1.

• A non-electrolyte neither dissociates nor associates, so i = 1.

Working out i for each solution:

• (A) Aqueous urea: non-electrolyte, does not dissociate: 1 particle, i = 1.

• (B) Aqueous K₂SO₄: \( \ce{K2SO4} \) → \( \ce{2K^{+}} \) + \( \ce{SO4^{2-}} \): 2 + 1 = 3 ions, i = 3.

• (C) Aqueous NaCl: NaCl → \( \ce{Na^{+}} \) + \( \ce{Cl^{-}} \): 1 + 1 = 2 ions, i = 2.

• (D) Ethanoic acid in benzene: in a non-polar solvent the acid molecules dimerise through hydrogen bonding, so two molecules become one particle, i ≈ 1/2.

      \( \displaystyle \ce{2CH3COOH <=> (CH3COOH)2} \)

Shortcut: Ethanoic acid in benzene is the only associated solute, so it must come first, and \( \ce{K2SO4} \) gives the most ions, so it must come last. Only Option (3) starts with D and ends with B.

NCERT (2026-27 edition):
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. …”

“In 1880 van’t Hoff introduced a factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. This factor i is defined as:

    \( \displaystyle i=\frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}} \)”