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 15). "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?
Answer: D. Freezing-point depression is a colligative property — it depends only on the number of solute particles, not their identity (NCERT Class 12 Chemistry Chapter 1, page 15).
Why A is wrong: A is wrong because viscosity depends on the nature of solute–solvent interactions, not solely on particle count.
Why B is wrong: B is wrong because electrical conductivity depends on the identity and charge of ions, not just their number.
Why C is wrong: C is wrong because optical rotation depends on the specific molecular structure (chirality) of the solute.
The cryoscopic constant K_f of water is approximately:
Answer: C. K_f for water is 1.86 K·kg/mol, a standard constant used in freezing-point depression calculations (NCERT Class 12 Chemistry Chapter 1).
Why A is wrong: A is wrong because 0.52 K·kg/mol is the ebullioscopic constant K_b of water, not the cryoscopic constant K_f.
Why B is wrong: B is wrong because 2.53 K·kg/mol is the ebullioscopic constant of benzene, not a constant of water.
Why D is wrong: D is wrong because 0.186 is a decimal-place error (K_f = 1.86, not 0.186).
The Van't Hoff factor i for CaCl₂, assuming complete dissociation, is:
Answer: B. CaCl₂ → Ca²⁺ + 2Cl⁻ gives 3 particles per formula unit, so i = 3 (NCERT Class 12 Chemistry Chapter 1).
Why A is wrong: A is wrong because i = 1 applies only to non-electrolytes. CaCl₂ is an ionic compound that dissociates.
Why C is wrong: C is wrong because i = 2 would apply to a 1:1 electrolyte like NaCl. CaCl₂ yields 3 ions, not 2.
Why D is wrong: D is wrong because CaCl₂ produces 3 ions (one Ca²⁺ and two Cl⁻), not 4. Counting 4 suggests incorrectly splitting Ca²⁺ further.
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)
Answer: A. Moles of urea = 6.0/60 = 0.10 mol. Molality = 0.10/0.500 = 0.20 mol/kg. ΔT_f = K_f × m = 1.86 × 0.20 = 0.372 K.
Why B is wrong: B is wrong because 0.186 K results from using 0.10 mol/kg as molality — this error comes from dividing by 1 kg instead of the actual 0.500 kg of solvent.
Why C is wrong: C is wrong because 1.86 K equals the K_f constant itself, suggesting the molality was taken as 1 mol/kg instead of calculating it correctly.
Why D is wrong: D is wrong because 0.093 K results from halving the correct answer, likely from an error in calculating molality (using 1 kg instead of 0.5 kg and then halving again).
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:
Answer: A. By Raoult's law: p = p°_A·x_A + p°_B·x_B = 300 × 0.4 + 100 × 0.6 = 120 + 60 = 180 mmHg.
Why B is wrong: B is wrong because 200 mmHg is the simple average of 300 and 100. Raoult's law requires mole-fraction weighting, not arithmetic averaging (trap: treating mole fractions as equal).
Why C is wrong: C is wrong because 160 mmHg results from swapping the mole fractions: using x_A = 0.6 and x_B = 0.4 instead of the given values.
Why D is wrong: D is wrong because 220 mmHg could arise from incorrectly using mass fractions or adding an extra term. Raoult's law gives exactly p°_A·x_A + p°_B·x_B.
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)
Answer: D. π = (W/M)(RT/V). Rearranging: M = WRT/(πV) = 3.0 × 0.0821 × 300 / (2.46 × 0.250) = 73.89/0.615 ≈ 120 g/mol.
Why A is wrong: A is wrong because 60 g/mol results from using V = 0.500 L (doubling the volume) instead of the correct 0.250 L.
Why B is wrong: B is wrong because 240 g/mol results from halving the molarity (perhaps using 125 mL or an extra factor of 2), which doubles the calculated molar mass.
Why C is wrong: C is wrong because 30 g/mol results from using V = 1 L instead of 0.250 L — a failure to convert mL to L correctly.
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)
Answer: B. NaCl dissociates completely: i = 2. ΔT_b = i × K_b × m = 2 × 0.52 × 0.1 = 0.104 K. Two steps: determine i from dissociation, then apply the corrected formula.
Why A is wrong: A is wrong because 0.052 K comes from omitting the Van't Hoff factor (i = 1 instead of 2). NaCl is an electrolyte — you must multiply by i (trap: forgetting Van't Hoff factor for ionic compounds).
Why C is wrong: C is wrong because 0.156 K results from using i = 3, which applies to CaCl₂ or K₂SO₄, not NaCl. NaCl gives only 2 ions (Na⁺ + Cl⁻).
Why D is wrong: D is wrong because 0.520 K is K_b × 1 × 1: it drops both the Van't Hoff factor and the given molality (m = 1.0 instead of 0.1).
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?
Answer: C. C is correct. Relative lowering of vapour pressure equals the solute mole fraction: (p° − p)/p° = (25.0 − 23.4)/25.0 = 1.6/25.0 = 0.064. Moles of solvent = 90/18 = 5.00. Then x_solute = n/(n + 5.00) = 0.064, so n = 0.064 × 5.00/(1 − 0.064) = 0.320/0.936 = 0.342 mol, and M = 10/0.342 = 29.2 ≈ 29 g/mol. Watch the two easy slips in that line: 0.064 × 5.00 is 0.320, and 0.320/0.936 is 0.342 — it is the 0.342 that is the number of moles, not an intermediate.
Why A is wrong: A is wrong: 170 g/mol does not fit the data. Check by substitution: 10/170 = 0.0588 mol of solute gives x_solute = 0.0588/5.059 = 0.0116 and a predicted p = 25.0 × (1 − 0.0116) = 24.7 mmHg, not the measured 23.4 mmHg.
Why B is wrong: B is wrong: by the same check, 10/34 = 0.294 mol gives x_solute = 0.294/5.294 = 0.0556 and p = 25.0 × (1 − 0.0556) = 23.6 mmHg, not 23.4 mmHg. The measured lowering needs x_solute = 0.064.
Why D is wrong: D is wrong, but only just, and the reason is worth knowing: 31 g/mol comes from the dilute approximation x_solute ≈ n_solute/n_solvent, which drops n_solute from the denominator and gives n = 0.064 × 5.00 = 0.320 mol and M = 10/0.320 = 31.3. At x = 0.064 the solution is not dilute enough for that: the exact form gives 29 g/mol, about 7% lower.
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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.)
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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. 15These 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 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
For NaCl i ≈ 2, CaCl₂ i ≈ 3, K₂SO₄ i ≈ 3. Multiply colligative formulas by i.
More in Solutions: 2 exam traps and mistakes · 8 formulas · 2 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
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
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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