The trap that costs marks here: you see "NaCl solution" in the stem, calculate ΔT_f perfectly — and forget to multiply by the Van't Hoff factor i. That single omission flips your answer to a distractor. Ionic solutes dissociate; non-electrolyte formulas don't account for the extra particles.
What colligative properties actually are. Four solution properties depend only on the number of solute particles, not their identity: relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, and osmotic pressure (NCERT Class 12 Chemistry Chapter 1, page 14). "Colligative" literally means "depending on collection" — count particles, not chemical nature.
The four formulas and when each fires.
All four use molality or mole fraction — temperature-independent concentration units (except osmotic pressure, which uses molarity).
The Van't Hoff correction. For electrolytes, multiply every colligative formula by i. NaCl → Na⁺ + Cl⁻, so i ≈ 2. CaCl₂ → Ca²⁺ + 2Cl⁻, so i ≈ 3. K₂SO₄ → 2K⁺ + SO₄²⁻, so i ≈ 3. Forgetting i is the single highest-frequency error in this topic.
The mole-fraction trap. Raoult's law requires mole fractions. When a problem gives you masses of two liquids, you must convert to moles first. Substituting mass fraction directly produces a distractor — and NTA knows this.
Watch-out: if the stem says "0.1 m NaCl," your effective molality is 0.1 × 2 = 0.2 m for colligative calculations.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following properties of a solution depends on the number of solute particles and NOT on their chemical nature?
The cryoscopic constant K_f of water is approximately:
The Van't Hoff factor *i* for CaCl₂, assuming complete dissociation, is:
A solution contains 6.0 g of urea (M = 60 g/mol) dissolved in 500 g of water. What is the freezing-point depression? (K_f for water = 1.86 K·kg/mol; urea is a non-electrolyte)
Two liquids A and B form an ideal solution. Pure vapour pressures: p°_A = 300 mmHg, p°_B = 100 mmHg. If x_A = 0.4, the total vapour pressure of the solution is:
The osmotic pressure of a solution containing 3.0 g of a solute (molar mass M) in 250 mL at 300 K is 2.46 atm. What is the molar mass of the solute? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)
0.1 mol of NaCl is dissolved in 1 kg of water. Assuming complete dissociation, the boiling-point elevation is: (K_b for water = 0.52 K·kg/mol)
A solution of a non-volatile solute has a vapour pressure of 23.4 mmHg. The vapour pressure of pure solvent is 25.0 mmHg. If the solvent's molar mass is 18 g/mol and the solution contains 90 g of solvent, what is the molar mass of the solute if 10 g of solute is present?
Pattern: Colligative properties calculation with Van't Hoff factor (pattern: calculate ΔT_f for an electrolyte).
Given
- Solute: CaCl₂, mass = 11.1 g, molar mass = 111 g/mol - Solvent: water, mass = 500 g = 0.500 kg - K_f for water = 1.86 K·kg/mol - Assume complete dissociation
Required
Find the freezing-point depression ΔT_f.
Concept
Freezing-point depression for an electrolyte requires the Van't Hoff correction: ΔT_f = i · K_f · m. CaCl₂ dissociates into Ca²⁺ + 2Cl⁻, giving i = 3.
Formula
ΔT_f = i · K_f · m, where m = n_solute / kg_solvent
Substitution
- n(CaCl₂) = 11.1 / 111 = 0.100 mol - m = 0.100 / 0.500 = 0.200 mol/kg - i = 3 (CaCl₂ → Ca²⁺ + 2Cl⁻) - ΔT_f = 3 × 1.86 × 0.200
Calculation
ΔT_f = 3 × 1.86 × 0.200 = 3 × 0.372 = 1.116 K Note: the integer 3 (Van't Hoff factor from ion count) is an exact counting number and does not limit significant figures. The calculation is limited to 3 significant figures by K_f (1.86) and m (0.200).
Final answer
ΔT_f = 1.12 K (3 significant figures) The new freezing point of the solution = 0.00 − 1.12 = −1.12 °C.
Common trap
Forgetting the Van't Hoff factor for CaCl₂ gives ΔT_f = 1.86 × 0.200 = 0.372 K — exactly one-third of the correct answer. This is a high-frequency distractor in NEET papers. Always check: is the solute ionic? If yes, determine *i* before calculating.
Similar NEET-style question
"Calculate the boiling-point elevation when 5.85 g of NaCl (M = 58.5 g/mol) is dissolved in 250 g of water. K_b = 0.52 K·kg/mol. Assume complete dissociation." (Answer approach: n = 0.100 mol, m = 0.400 mol/kg, i = 2, ΔT_b = 2 × 0.52 × 0.400 = 0.416 K.) ---
Properties depending only on number of solute particles, not their nature: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure.
-- NCERT Class 12 Chemistry, Ch. 1, p. 14Molal concentration: moles of solute per kg of solvent. Temperature-independent.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m | molality | mol/kg |
| n | moles solute | mol |
Molar concentration: moles of solute per litre of solution.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | molarity | mol/L |
| n | moles solute | mol |
| V | solution volume | L |
Solute raises boiling point. K_b is ebullioscopic constant of solvent (water: 0.52 K kg/mol).
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔT_b | BP elevation | K |
| K_b | ebullioscopic constant | K kg/mol |
| m | molality | mol/kg |
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 |
Pressure required to prevent osmosis. C in mol/L; T in K. Used for high-molar-mass biomolecules.
| Symbol | Quantity | SI Unit |
|---|---|---|
| π | osmotic pressure | Pa |
| C | molarity | mol/L |
| R | gas constant | J/mol/K |
| T | temp | K |
Total vapor pressure of ideal solution = sum of mole-fraction-weighted vapor pressures of components.
| Symbol | Quantity | SI Unit |
|---|---|---|
| p | total vapor pressure | Pa |
| p_i° | pure component vp | Pa |
| x_i | mole fraction | - |
For non-volatile solute: relative lowering of VP equals mole fraction of solute.
| Symbol | Quantity | SI Unit |
|---|---|---|
| p | solution vp | Pa |
| p° | pure solvent vp | Pa |
| x_solute | mole fraction | - |
Correction factor for electrolytes. NaCl: i≈2; CaCl₂: i≈3. Multiply colligative formula by i.
| Symbol | Quantity | SI Unit |
|---|---|---|
| i | Van't Hoff factor | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student uses mass fraction (w₁/total mass) where mole fraction (n₁/total moles) is required.
Question gives masses or molar masses and asks about Raoult's law or vapor pressure.
Raoult's law uses MOLE fractions, not mass fractions. Convert mass to moles first using molar mass.
Category: Similar Terms
Student uses non-electrolyte colligative formula for ionic compound. NaCl: i ≈ 2; CaCl₂: i ≈ 3.
Question gives an ionic compound (NaCl, CaCl₂, K₂SO₄) and asks for colligative property.
For electrolytes, multiply colligative formula by Van't Hoff factor i. NaCl → Na⁺ + Cl⁻ (i=2). CaCl₂ → Ca²⁺ + 2Cl⁻ (i=3). K₂SO₄ → 2K⁺ + SO₄²⁻ (i=3).
Root cause: formula misuse
Convert mass to moles first using molar mass. Raoult uses MOLE fractions.
Root cause: formula misuse
For NaCl i ≈ 2, CaCl₂ i ≈ 3, K₂SO₄ i ≈ 3. Multiply colligative formulas by i.
9 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
Which of the following aqueous solution will exhibit highest boiling point?
The IUPAC name of an element with atomic number 119 is
The structures of beryllium chloride in solid state and vapour phase, are :
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
forgets i factor electrolytes
Uses non-electrolyte formula for ionic solute
forgets conversion of units
Mixes Pa with atm; L with m^3
uses mass fractions not mole
Substitutes mass fraction for x
Test yourself on this topic with real past-paper questions:
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