Which one of the following solutions will exhibit smallest freezing point depression?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(0.1\, m\, C_6H_{12}O_6\)
Depression of freezing point is a colligative property: it depends on the number of solute particles, not on their nature. For an electrolyte the van’t Hoff factor i counts the particles formed by one formula unit:
\( \Delta T_f = i\,K_f\,m \)
All four solutions are 0.1 m in the same solvent, so ΔTf is proportional to i. Writing the dissociation of each and counting the particles:
\( \ce{K2SO4} \) → \( \ce{2K^{+}} \) + \( \ce{SO4^{2-}} \): 2 + 1 = 3 ions, i = 3
\( \ce{C6H12O6} \) (glucose) does not dissociate: 1 particle, i = 1
\( \ce{Al2(SO4)3} \) → \( \ce{2Al^{3+}} \) + \( \ce{3SO4^{2-}} \): 2 + 3 = 5 ions, i = 5
KCl → \( \ce{K^{+}} \) + \( \ce{Cl^{-}} \): 1 + 1 = 2 ions, i = 2
The smallest i, and so the smallest depression, belongs to the non-electrolyte glucose.
Why the other options are incorrect
Option (1) \(0.1\, m\, K_2SO_4\) — Incorrect. i = 3, so its depression is three times that of glucose.
Option (3) \(0.1\, m\, Al_2(SO_4)_3\) — Incorrect. i = 5, the largest depression of the four.
Option (4) \(0.1\, m\, KCl\) — Incorrect. i = 2, twice the depression of glucose.
NCERT (2026-27 edition):
Abnormal Molar Masses – “… Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows: Relative lowering of vapour pressure of solvent, \(\frac{p_1^\circ - p_1}{p_1^\circ} = i\,\frac{n_2}{n_1}\) Elevation of Boiling point, \(\Delta T_b = i\,K_b\,m\) Depression of Freezing point, \(\Delta T_f = i\,K_f\,m\) Osmotic pressure of solution, \(\pi = i\,n_2\,R\,T/V\)”