Formula
Freezing point depression
ΔT_f = K_f × m, where K_f is cryoscopic constant (water: 1.86 K·kg/mol). Used to determine molar mass of solute.
-- NCERT Class 12 Chemistry, Ch. 1, p. 18Depression Freezing Point
8 MCQs3 revision cards9-step worked example
Source: NCERT SolutionsPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 25 Sep 2026
Depression Freezing Point, explained for NEET
Depression of Freezing Point
Here is the trap that costs marks on this topic: you see NaCl or CaCl₂ in the stem, calculate ΔT_f using the plain formula, and forget to multiply by the Van't Hoff factor i. That single omission produces a distractor that NTA places among the options every time this topic appears.
The core idea. A non-volatile solute dissolved in a solvent lowers the solvent's freezing point. The depression is a colligative property — it depends on the number of solute particles, not their identity. NCERT Class 12 Chemistry Chapter 1 (page 20) gives the relationship:
ΔT_f = K_f × m
where K_f is the cryoscopic constant of the solvent (water: 1.86 K·kg/mol) and m is molality (moles of solute per kg of solvent — not per kg of solution).
For electrolytes, each formula unit dissociates into ions, increasing the effective particle count. The corrected form is:
ΔT_f = i × K_f × m
where i is the Van't Hoff factor. NaCl → Na⁺ + Cl⁻ gives i ≈ 2. CaCl₂ → Ca²⁺ + 2 Cl⁻ gives i ≈ 3. Forgetting i gives you exactly half (or one-third) of the correct answer — and that value will be sitting in the options waiting for you.
Molar mass from ΔT_f. A common NEET pattern: you observe a freezing-point depression, know K_f and the mass of solute, and back-calculate molar mass. Rearrange: M₂ = (K_f × w₂ × 1000) / (ΔT_f × w₁), where w₂ is mass of solute in grams and w₁ is mass of solvent in grams. For electrolytes, replace K_f with i × K_f (or equivalently divide the calculated M₂ by i).
Watch-out: molality uses mass of solvent, not mass of solution. If the problem gives total solution mass, subtract the solute mass first.
Can you answer these Depression Freezing Point MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The freezing-point depression of a solution is a colligative property because it depends on:
Show answer and why every option is right or wrong
Answer: C. Colligative properties depend on the number (concentration) of solute particles, not their chemical identity. This is the defining feature stated in NCERT Class 12 Chemistry Chapter 1.
Why A is wrong: A is wrong because colligative properties are independent of the nature (identity) of solute particles — only particle count matters.
Why B is wrong: B is wrong because molar mass determines how many moles a given mass produces, but the property itself depends on particle count, not molar mass directly.
Why D is wrong: D is wrong because the boiling point of the pure solvent is a solvent property unrelated to the definition of colligative.
MCQ 2Easy RecallPractice
The SI unit of the cryoscopic constant K_f is:
Show answer and why every option is right or wrong
Answer: B. From ΔT_f = K_f × m, K_f = ΔT_f / m. Units: K ÷ (mol/kg) = K·kg/mol. NCERT Class 12 Chemistry Chapter 1, page 20.
Why A is wrong: A is wrong because K·mol/kg inverts the denominator — dimensionally this would make ΔT_f have units of K·mol²/kg², which is nonsensical.
Why C is wrong: C is wrong because while °C and K have the same magnitude for differences, the standard SI expression uses K, and the denominator must be mol/kg (molality), not just mol.
Why D is wrong: D is wrong because K/mol omits the kg term. K_f must cancel the kg in molality's mol/kg to yield K for ΔT_f.
MCQ 3Easy RecallPractice
Which of the following aqueous solutions will show the greatest depression in freezing point? (Assume complete dissociation and equal molality for all.)
Show answer and why every option is right or wrong
Answer: B. ΔT_f = i × K_f × m. At equal molality, the solution with the highest i wins. CaCl₂ → Ca²⁺ + 2 Cl⁻ gives i = 3, which is the largest among the options. NCERT Class 12 Chemistry Chapter 1.
Why A is wrong: A is wrong because glucose is a non-electrolyte (i = 1), giving ΔT_f = 1 × K_f × 0.1, the smallest depression among the four.
Why C is wrong: C is wrong because NaCl gives i ≈ 2 (Na⁺ + Cl⁻), which is less than CaCl₂'s i ≈ 3. (Trap: forgetting to compare Van't Hoff factors.)
Why D is wrong: D is wrong because urea is a non-electrolyte (i = 1), identical effective particle count to glucose, giving the same low ΔT_f.
MCQ 4Direct ApplicationPractice
5.85 g of NaCl (molar mass = 58.5 g/mol) is dissolved in 500 g of water. If K_f for water is 1.86 K·kg/mol and NaCl dissociates completely, the depression in freezing point is:
Show answer and why every option is right or wrong
Answer: D. Moles of NaCl = 5.85/58.5 = 0.1 mol. Molality = 0.1/0.500 = 0.2 mol/kg. For NaCl, i = 2. ΔT_f = i × K_f × m = 2 × 1.86 × 0.2 = 0.744 K. NCERT Class 12 Chemistry Chapter 1.
Why A is wrong: A is wrong because 0.372 K is the value obtained WITHOUT the Van't Hoff factor (ΔT_f = 1.86 × 0.2 = 0.372). This is the classic trap of forgetting i for an ionic compound.
Why B is wrong: B is wrong because 3.72 K uses i = 2 but an incorrect molality of 1 mol/kg (dividing moles by mass in grams instead of kilograms: 0.1/0.5 should give 0.2, not 1.0 from 0.1/0.1).
Why C is wrong: C is wrong because 1.86 K would require m = 1 mol/kg with i = 1, or m = 0.5 with i = 2. Neither matches the given data — this likely results from a molality calculation error (using 0.1/0.1 = 1 instead of 0.1/0.5 = 0.2).
MCQ 5Direct ApplicationPractice
3.0 g of a non-electrolyte solute is dissolved in 100 g of water. The freezing point of the solution is found to be −0.93 °C. If K_f for water is 1.86 K·kg/mol, the molar mass of the solute is:
Show answer and why every option is right or wrong
Answer: D. ΔT_f = 0 − (−0.93) = 0.93 K. Using M₂ = (K_f × w₂ × 1000)/(ΔT_f × w₁) = (1.86 × 3.0 × 1000)/(0.93 × 100) = 5580/93 = 60 g/mol. Non-electrolyte so i = 1. NCERT Class 12 Chemistry Chapter 1.
Why A is wrong: A is wrong because 30 g/mol results from doubling ΔT_f (using 1.86 instead of 0.93) or halving the formula — a calculation slip.
Why B is wrong: B is wrong because 120 g/mol results from using 50 g instead of 100 g for the solvent mass — an error in reading the given data.
Why C is wrong: C is wrong: 90 g/mol would mean 3.0/90 = 0.033 mol in 0.100 kg, m = 0.33 mol/kg and ΔT_f = 1.86 × 0.33 = 0.62 K, not the measured 0.93 K.
MCQ 6Direct ApplicationPractice
The freezing point of a 0.05 m aqueous solution of K₂SO₄ is (assume complete dissociation, K_f = 1.86 K·kg/mol):
Show answer and why every option is right or wrong
Answer: A. K₂SO₄ → 2 K⁺ + SO₄²⁻, so i = 3. ΔT_f = i × K_f × m = 3 × 1.86 × 0.05 = 0.279 K. Freezing point = 0 − 0.279 = −0.279 °C. NCERT Class 12 Chemistry Chapter 1.
Why B is wrong: B is wrong because −0.186 °C uses i = 2 (perhaps counting only one K⁺ and one SO₄²⁻, forgetting K₂SO₄ produces TWO potassium ions). Correct i = 3.
Why C is wrong: C is wrong because −0.093 °C uses i = 1 (treating K₂SO₄ as a non-electrolyte: 1 × 1.86 × 0.05 = 0.093). This is the Van't Hoff factor trap.
Why D is wrong: D is wrong because −0.372 °C uses i = 4, which would mean four ions per formula unit. K₂SO₄ produces 3 ions (2 K⁺ + 1 SO₄²⁻), not 4.
MCQ 7CalculationPractice
0.6 g of urea (molar mass 60 g/mol) and 0.585 g of NaCl (molar mass 58.5 g/mol) are dissolved together in 200 g of water. If K_f = 1.86 K·kg/mol and NaCl dissociates completely, the freezing-point depression is:
Show answer and why every option is right or wrong
Answer: C. Moles urea = 0.6/60 = 0.01 mol (i = 1). Moles NaCl = 0.585/58.5 = 0.01 mol (i = 2, effective particles = 0.02 mol). Total effective moles = 0.01 + 0.02 = 0.03. Molality (effective) = 0.03/0.200 = 0.15 mol/kg. ΔT_f = K_f × m_eff = 1.86 × 0.15 = 0.279 K.
Why A is wrong: A is wrong because 0.093 K results from using only the urea contribution (0.01/0.2 × 1.86 = 0.093) and ignoring NaCl entirely.
Why B is wrong: B is wrong because 0.186 K results from treating both solutes as non-electrolytes (i = 1 for both), giving total moles = 0.02 and m = 0.1 mol/kg: 1.86 × 0.1 = 0.186. Trap: forgetting i = 2 for NaCl.
Why D is wrong: D is wrong because 0.372 K results from using i = 2 for BOTH solutes (including urea, which is a non-electrolyte). Urea does not dissociate; i = 1.
MCQ 8CalculationPractice
An aqueous solution of a substance X freezes at −0.558 °C. When 1.0 g of X is dissolved in 100 g of water, K_f = 1.86 K·kg/mol. The solution is found to have Van't Hoff factor i = 2. The molar mass of X (as an undissociated formula unit) is:
Show answer and why every option is right or wrong
Answer: A. ΔT_f = 0.558 K. Using ΔT_f = i × K_f × (w₂ × 1000)/(M₂ × w₁), rearrange: M₂ = i × K_f × w₂ × 1000 / (ΔT_f × w₁) = 2 × 1.86 × 1.0 × 1000 / (0.558 × 100) = 3720 / 55.8 ≈ 66.7 g/mol.
Why B is wrong: B is wrong because 33.3 g/mol results from omitting the Van't Hoff factor i in the numerator (using 1 × 1.86 instead of 2 × 1.86), giving exactly half the correct answer. This is the i-factor trap.
Why C is wrong: C is wrong because 100 g/mol results from using i = 3, as for a salt that gives three ions, when the question states i = 2: 3 × 1.86 × 1.0 × 1000 / (0.558 × 100) = 100.
Why D is wrong: D is wrong because 133.3 g/mol counts the Van't Hoff factor twice (i² = 4 in the numerator): 4 × 1.86 × 1.0 × 1000 / (0.558 × 100) = 133.3, double the correct answer.
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Depression Freezing Point: quick recall before you leave
How do you solve a Depression Freezing Point question? A worked example
- 1
Given
• 11.1 g of CaCl₂ (molar mass = 111 g/mol) dissolved in 500 g of water• K_f for water = 1.86 K·kg/mol• CaCl₂ dissociates completely
- 2
Required
Depression of freezing point, ΔT_f
- 3
Concept
Freezing-point depression is a colligative property. For an electrolyte, the effective particle count is higher than the formula-unit count, so we must include the Van't Hoff factor i. CaCl₂ → Ca²⁺ + 2 Cl⁻ produces 3 ions per formula unit, so i = 3 (NCERT Class 12 Chemistry Chapter 1, page 20).
- 4
Formula
ΔT_f = i × K_f × m, where m = (moles of solute) / (kg of solvent)
- 5
Substitution
• Moles of CaCl₂ = 11.1 g / 111 g/mol = 0.1 mol• Mass of solvent = 500 g = 0.500 kg• m = 0.1 / 0.500 = 0.2 mol/kg• ΔT_f = 3 × 1.86 × 0.2
- 6
Calculation
ΔT_f = 3 × 1.86 × 0.2 = 3 × 0.372 = 1.116 K
Note on exact values: the integer 3 (ion count) and the division 11.1/111 = 0.1 (exact by construction of the problem) are counting/exact quantities and do not limit significant figures. K_f = 1.86 (3 sig figs) governs precision. - 7
Final answer
ΔT_f = 1.12 K (3 significant figures, governed by K_f = 1.86)
Freezing point of solution = 0 − 1.12 = −1.12 °C - 8
Common trap
Without the Van't Hoff factor: ΔT_f = 1.86 × 0.2 = 0.372 K — exactly one-third of the correct answer. This is the value NTA places as a distractor. If your answer is suspiciously small for an ionic compound, check whether you included i.
- 9
Similar NEET-style question
*What is the freezing point of a solution containing 5.85 g of NaCl (M = 58.5 g/mol) in 250 g of water? (K_f = 1.86 K·kg/mol, complete dissociation)*
Approach: moles = 0.1, m = 0.1/0.250 = 0.4 mol/kg, i = 2 for NaCl. ΔT_f = 2 × 1.86 × 0.4 = 1.488 K. Freezing point = −1.49 °C.
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What to remember before solving Depression Freezing Point questions
Which Depression Freezing Point formulas do you need for NEET?
Freezing-point depression
Solute lowers freezing point. K_f is cryoscopic constant of solvent (water: 1.86 K kg/mol). Used for molar mass determination.
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔT_f | FP depression | K |
| K_f | cryoscopic constant | K kg/mol |
| m | molality | mol/kg |
Valid when
- Dilute solution
- Non-electrolyte (else multiply by i)
More in Solutions: 4 exam traps and mistakes · 7 formulas · 2 question patterns from its other lessons.
Depression Freezing Point questions from past NEET papers
1 question from NEET 2020. Answers verified against NTA official keys.
NEET 2020
10.60 K
20.20 K
30.80 K
40.40 K
NTA Answer: Option 4(final)
How does NEET ask about Depression Freezing Point?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Calculate ΔT_f or ΔT_b given molality. Or determine molar mass from observed colligative property.
Multi StepMedium
Common distractors
forgets i factor electrolytes
Uses non-electrolyte formula for ionic solute
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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