Which of the following 0.1 M aqueous solutions will have the lowest freezing point?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → K2SO4
Depression of freezing point is a colligative property: it depends on the number of solute particles. For a solute that dissociates, the van’t Hoff factor i counts the particles formed from one formula unit:
\( \displaystyle \Delta T_f=i\,K_f\,m \)
All four solutions are 0.1 M in water, so Kf and the concentration are the same, and ΔTf depends only on i. The largest i gives the largest depression, and so the lowest freezing point.
\( \displaystyle \ce{K2SO4 -> 2K^+ + SO4^2-} \)
2 + 1 = 3 ions, so i = 3, the largest of the four. \( \ce{K2SO4} \) has the lowest freezing point.
Why the other options are incorrect
Option (2) NaCl — Incorrect. \( \ce{NaCl -> Na^+ + Cl^-} \) gives 2 ions, so i = 2: a smaller depression than \( \ce{K2SO4} \).
Option (3) Urea — Incorrect. Urea is a non-electrolyte; it does not dissociate, so i = 1 and its freezing point is among the highest of the four.
Option (4) Glucose — Incorrect. Glucose is also a non-electrolyte with i = 1, so it lowers the freezing point by the same small amount as urea.
NCERT (2026-27 edition):
1.7 Abnormal Molar Masses – “We know that ionic compounds when dissolved in water dissociate into cations and anions. … As expected, the value of i gets close to 3 for K2SO4.”