Question 21 markMCQSection A

Which of the following solutions will have the lowest freezing point in water?

Solutions
Depression in freezing point and van't Hoff factor

Options

(A)0.1 M Glucose
(B)0.1 M CaCl₂
(C)0.1 M KCl
(D)0.1 M Urea

Official answer

Correct option: (B) — 0.1 M CaCl₂.


Key idea: ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m; lowest freezing point means the largest depression ΔTf, i.e. the largest van't Hoff factor i.


  • CaCl2→Ca2++2Cl−CaCl_2 \rightarrow Ca^{2+} + 2Cl^-, so i=3i = 3 (highest).
  • KCl→K++Cl−KCl \rightarrow K^+ + Cl^-, i=2i = 2.
  • Glucose and Urea are non-electrolytes, i=1i = 1.

Since all are 0.1 M, CaCl₂ gives the greatest ΔTf and therefore the lowest freezing point.

Answer keywords:depression in freezing pointcolligative propertyvan't Hoff factorCaCl₂ i = 3electrolyte dissociationΔTf = i·Kf·mnumber of particles

Marking scheme

  1. 1.1 mark: correctly selecting option (B) CaCl₂.
  2. 2.Justification (i=3i = 3, largest ΔTf\Delta T_f) is the accepted reasoning.

Hint

Lowest freezing point = biggest ΔTf\Delta T_f = biggest van't Hoff factor i. Count the ions.

Quick oral answer

0.1 M CaCl₂ has the lowest freezing point because it dissociates into three ions, giving a van't Hoff factor of three — the greatest number of particles, hence the greatest depression of freezing point.

Common mistakes

  1. 1.Comparing molarity only and ignoring the van't Hoff factor for electrolytes.
  2. 2.Assuming glucose or urea depress the freezing point most because they are 'organic' — non-electrolytes actually depress it least.
  3. 3.Forgetting CaCl₂ gives three ions (one Ca²⁺ and two Cl⁻), taking i=2i = 2 by mistake.

Analysis and explanation

Concept:

Depression of freezing point is a colligative property: ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m. It depends on the number of solute particles, captured by the van't Hoff factor i.


Particle count at 0.1 M:


  • CaCl₂ → 3 ions ⇒ i=3i = 3 ⇒ effective 0.3 mol particles.
  • KCl → 2 ions ⇒ i=2i = 2 ⇒ 0.2 mol particles.
  • Glucose (non-ionic) ⇒ i=1i = 1 ⇒ 0.1 mol particles.
  • Urea (non-ionic) ⇒ i=1i = 1 ⇒ 0.1 mol particles.

Most particles ⇒ largest ΔTf ⇒ lowest freezing point ⇒ CaCl₂ (B).


Why the distractors are wrong:


  • (A) Glucose and (D) Urea do not dissociate (i=1i = 1); they cause the smallest depression.
  • (C) KCl dissociates into only 2 ions (i=2i = 2), so its depression is less than CaCl₂'s.

Exam trap: Never compare molarities alone — always multiply by i for electrolytes.

Interesting facts

CaCl₂ is spread on icy roads in winter precisely because its i=3i = 3 lets it melt ice at lower temperatures (down to about −29 °C) than common NaCl.

The van't Hoff factor earned Jacobus van't Hoff the first-ever Nobel Prize in Chemistry (1901) for his work on solution theory and osmotic pressure.

Question details

Q2 of 37 questions

Board
CBSE
Class
Class 12
Subject
Chemistry
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Solutions
Topic
Depression in freezing point and van't Hoff factor
Bloom's level
apply
Syllabus topic
Unit 2: Solutions — Colligative Properties, Depression of Freezing Point and van't Hoff Factor
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why does CaCl₂ lower the freezing point more than KCl at the same molarity?

Depression of freezing point depends on the number of dissolved particles, not their identity. CaCl₂ releases three ions per formula unit (one Ca²⁺ and two Cl⁻, i=3i = 3) whereas KCl releases only two (K⁺ and Cl⁻, i=2i = 2). More particles from CaCl₂ mean a larger ΔTf\Delta T_f, so its solution freezes at a lower temperature.

Do glucose and urea affect the freezing point at all?

Yes, but the least of the four. Both are non-electrolytes that dissolve as single, undissociated molecules (i=1i = 1), so a 0.1 M solution supplies only 0.1 mol of particles per litre. They still depress the freezing point, just far less than the electrolytes CaCl₂ and KCl.