Q2
1 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.

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

Marking Scheme

  • 11 mark: correctly selecting option (B) CaCl₂.
  • 2Justification (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.

Analysis & 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.

Common Mistakes

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

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.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

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.