Here is the trap that costs marks on this topic: you see "NaCl solution" in the stem, calculate ΔT_f or ΔT_b using the standard colligative formula, get a clean number — and pick the wrong option because you forgot to multiply by the Van't Hoff factor i.
The Van't Hoff factor corrects colligative property formulas for solutes that dissociate (electrolytes) or associate in solution. NCERT Class 12 Chemistry Chapter 1, page 24 defines it as:
i = (observed colligative property) / (calculated colligative property for non-electrolyte)
For complete dissociation: NaCl → Na⁺ + Cl⁻ gives i = 2. CaCl₂ → Ca²⁺ + 2 Cl⁻ gives i = 3. K₂SO₄ → 2 K⁺ + SO₄²⁻ gives i = 3. Acetic acid dimerising in benzene gives i < 1 (association).
Every colligative formula gets multiplied by i when the solute is not a non-electrolyte:
Abnormal molar mass is the direct consequence. If you measure a colligative property and back-calculate molar mass without accounting for i, you get a molar mass that is too low (for dissociating solutes) or too high (for associating solutes). The relationship is:
i = (normal molar mass) / (abnormal molar mass)
The connection between i and degree of dissociation α for a solute producing n ions:
i = 1 + (n − 1)α
This lets NEET frame questions in both directions: given α, find i; or given observed vs. expected colligative property, find α.
Watch-out: When a question says "assuming complete dissociation," i equals the total number of ions per formula unit — no partial α needed.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The Van't Hoff factor i for a non-electrolyte dissolved in water is:
Answer: D. A non-electrolyte neither dissociates nor associates. The observed colligative property equals the calculated value, so i = observed/calculated = 1 (NCERT Class 12 Chemistry Chapter 1, page 24).
Why A is wrong: A is wrong because i = 0 would mean zero colligative effect. Non-electrolytes like glucose do show colligative properties — they just don't dissociate or associate.
Why B is wrong: B is wrong because i > 1 indicates dissociation (electrolyte behavior). A non-electrolyte has i = 1 exactly.
Why C is wrong: C is wrong because i = 2 applies to electrolytes that fully dissociate into 2 ions (e.g., NaCl), not to non-electrolytes.
For a solute that associates in solution, the Van't Hoff factor i is:
Answer: A. Association reduces the effective number of solute particles. Fewer particles means a smaller observed colligative property, so i = observed/calculated < 1. Example: acetic acid dimerising in benzene (NCERT Class 12 Chemistry Chapter 1, page 24).
Why B is wrong: B is wrong because i > 1 applies to dissociation (more particles than formula units). Association produces fewer particles, giving i < 1.
Why C is wrong: C is wrong because i = 1 applies to non-electrolytes that neither dissociate nor associate. Association reduces i below 1.
Why D is wrong: D is wrong because 'number of ions formed' describes dissociation of electrolytes. Association is the opposite process — molecules combine, reducing the particle count.
Assuming complete dissociation, the Van't Hoff factor for K₂SO₄ is:
Answer: B. K₂SO₄ → 2 K⁺ + SO₄²⁻ produces 3 ions per formula unit. With complete dissociation, i = 3 (NCERT Class 12 Chemistry Chapter 1, page 24).
Why A is wrong: A is wrong because i = 1 means no dissociation. K₂SO₄ is a strong electrolyte that dissociates into 3 ions.
Why C is wrong: C is wrong because i = 2 would mean only 2 particles form. K₂SO₄ gives 2 K⁺ ions and 1 SO₄²⁻ ion — that is 3 particles total (trap: forgetting Van't Hoff factor or miscounting ions).
Why D is wrong: D is wrong because i = 4 overcounts. There are only 3 ions per formula unit (2 K⁺ + 1 SO₄²⁻), not 4. The subscript '4' in SO₄²⁻ refers to oxygen atoms within one polyatomic ion, not separate ions.
0.1 m aqueous NaCl solution (assume complete dissociation, K_f for water = 1.86 K kg/mol). The freezing-point depression ΔT_f is:
Answer: A. NaCl → Na⁺ + Cl⁻, so i = 2. ΔT_f = i · K_f · m = 2 × 1.86 × 0.1 = 0.372 K. The Van't Hoff factor doubles the non-electrolyte value (NCERT Class 12 Chemistry Chapter 1, page 24).
Why B is wrong: B is wrong because 0.186 K = K_f × m = 1.86 × 0.1, which omits the Van't Hoff factor. NaCl is an electrolyte with i = 2; the colligative formula must be multiplied by i (trap: forgetting Van't Hoff factor for ionic compounds).
Why C is wrong: C is wrong because 0.558 K = 3 × 1.86 × 0.1, implying i = 3. NaCl produces only 2 ions (Na⁺ + Cl⁻), so i = 2, not 3.
Why D is wrong: D is wrong because 1.86 K = K_f × 1 = 1.86 × 1, which uses m = 1 instead of the given m = 0.1.
A 0.05 m aqueous solution of CaCl₂ (assume complete dissociation, K_b for water = 0.52 K kg/mol). The boiling-point elevation ΔT_b is:
Answer: C. CaCl₂ → Ca²⁺ + 2 Cl⁻ gives i = 3. ΔT_b = i · K_b · m = 3 × 0.52 × 0.05 = 0.078 K (NCERT Class 12 Chemistry Chapter 1, page 24).
Why A is wrong: A is wrong because 0.026 K = 0.52 × 0.05 = K_b × m, omitting the Van't Hoff factor entirely (trap: forgetting Van't Hoff factor for ionic compounds). CaCl₂ is an electrolyte with i = 3.
Why B is wrong: B is wrong because 0.052 K = 2 × 0.52 × 0.05, implying i = 2. CaCl₂ dissociates into 3 ions (Ca²⁺ + 2 Cl⁻), so i = 3, not 2. The subscript '2' in CaCl₂ refers to chloride ions, not the total ion count.
Why D is wrong: D is wrong because 0.156 K = 6 × 0.52 × 0.05, implying i = 6. CaCl₂ produces only 3 ions per formula unit, not 6.
The degree of dissociation of an electrolyte that produces 3 ions per formula unit is 0.8. Its Van't Hoff factor i is:
Answer: C. Using i = 1 + (n − 1)α with n = 3 and α = 0.8: i = 1 + (3 − 1)(0.8) = 1 + 1.6 = 2.6 (NCERT Class 12 Chemistry Chapter 1, page 24).
Why A is wrong: A is wrong because 1.8 = 1 + (n − 1)α with n = 2, not n = 3. The question specifies 3 ions per formula unit, so (n − 1) = 2, giving 1 + 2(0.8) = 2.6.
Why B is wrong: B is wrong because i = 3.0 assumes complete dissociation (α = 1). The question gives α = 0.8, so i = 1 + (3 − 1)(0.8) = 2.6, not the maximum value of 3.
Why D is wrong: D is wrong because 2.4 = 1 + (n − 1)α with α = 0.7 or a miscalculated (n − 1). With n = 3 and α = 0.8, the arithmetic gives 1 + 2(0.8) = 2.6.
A solute has a normal molar mass of 120 g/mol. When dissolved in water, its observed molar mass (from freezing-point depression) is 40 g/mol. How many ions does each formula unit produce on complete dissociation?
Answer: D. Step 1: i = normal molar mass / abnormal molar mass = 120/40 = 3. Step 2: Complete dissociation means α = 1, so i = n (number of ions). Therefore n = 3. The solute produces 3 ions per formula unit (NCERT Class 12 Chemistry Chapter 1, page 24).
Why A is wrong: A is wrong because n = 2 would give i = 2. But i = 120/40 = 3, so the solute produces 3 ions, not 2. Picking n = 2 likely comes from confusing the ratio with something else.
Why B is wrong: B is wrong because n = 6 would require abnormal molar mass = 120/6 = 20 g/mol. The question states 40 g/mol, giving i = 3.
Why C is wrong: C is wrong because n = 4 would require abnormal molar mass = 120/4 = 30 g/mol. The observed molar mass is 40 g/mol, giving i = 3, not 4.
An electrolyte AB₂ has a degree of dissociation α = 0.6 in water. Its Van't Hoff factor is i. If 0.2 m of this solute is dissolved in water (K_f = 1.86 K kg/mol), the freezing-point depression ΔT_f is:
Answer: B. Step 1: AB₂ → A²⁺ + 2 B⁻ gives n = 3 ions. Step 2: i = 1 + (n − 1)α = 1 + (2)(0.6) = 2.2. Step 3: ΔT_f = i · K_f · m = 2.2 × 1.86 × 0.2 = 0.8184 ≈ 0.818 K (NCERT Class 12 Chemistry Chapter 1, page 24).
Why A is wrong: A is wrong because 0.372 K = 1 × 1.86 × 0.2, using i = 1. AB₂ is an electrolyte — omitting the Van't Hoff factor ignores dissociation entirely (trap: forgetting Van't Hoff factor for ionic compounds).
Why C is wrong: C is wrong because 0.595 K ≈ 1.6 × 1.86 × 0.2, implying i = 1.6. This uses n = 2 (only 2 ions) in the formula i = 1 + (n − 1)α = 1 + (1)(0.6) = 1.6. AB₂ dissociates into 3 ions (n = 3), not 2.
Why D is wrong: D is wrong because 1.116 K = 3 × 1.86 × 0.2, using i = 3, which assumes complete dissociation (α = 1). The question gives α = 0.6, so i = 2.2, not 3.
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Pattern: Calculate ΔT_f for an ionic solute using the Van't Hoff factor (aligned with P.CHE.U05.COLLIGATIVE_PROPERTIES_CALC).
Given
• Mass of NaCl = 5.85 g• Molar mass of NaCl = 58.5 g/mol (exact, problem-defined)• Mass of solvent (water) = 500 g = 0.500 kg• K_f = 1.86 K kg/mol• Complete dissociation assumed
Required
ΔT_f = ?
Concept
Freezing-point depression for an electrolyte uses the Van't Hoff-corrected formula: ΔT_f = i · K_f · m. NaCl dissociates completely into Na⁺ and Cl⁻, giving i = 2.
Formula
ΔT_f = i · K_f · m, where m = (moles of solute) / (mass of solvent in kg)
Substitution
Moles of NaCl = 5.85 / 58.5 = 0.1 mol
Molality m = 0.1 / 0.500 = 0.2 mol/kg
ΔT_f = 2 × 1.86 × 0.2
Calculation
ΔT_f = 2 × 1.86 × 0.2 = 2 × 0.372 = 0.744 K
Note on exact values: The molar mass 58.5 g/mol, solvent mass 500 g, and the integer 2 (Van't Hoff factor for complete dissociation) are problem-defined exact values and do not limit significant figures in the final answer.
Final answer
ΔT_f = 0.744 K
Common trap
Without the Van't Hoff factor: ΔT_f = K_f · m = 1.86 × 0.2 = 0.372 K — exactly half the correct answer. This is the classic trap for ionic solutes: forgetting to multiply by i. NEET distractors routinely include this half-value as an option.
Similar NEET-style question
"0.1 m aqueous CaCl₂ (complete dissociation). K_f = 1.86 K kg/mol. Find ΔT_f." Answer: i = 3, ΔT_f = 3 × 1.86 × 0.1 = 0.558 K. The non-electrolyte trap answer would be 0.186 K.
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i = (observed colligative property) / (calculated value as if non-electrolyte). For ionic compounds dissociating into n ions: i ≈ n. Modify colligative formulas: ΔT_b = i K_b m, etc.
-- NCERT Class 12 Chemistry, Ch. 1, p. 24Correction 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 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 · 7 formulas · 3 question patterns from its other lessons.
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