Formula
Boiling point elevation
ΔT_b = K_b × m, where K_b is ebullioscopic constant of solvent (water: 0.52 K·kg/mol), m is molality.
-- NCERT Class 12 Chemistry, Ch. 1, p. 17Elevation Boiling Point
8 MCQs3 revision cards9-step worked example
Source: NCERT SolutionsPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 26 Sep 2026
Elevation Boiling Point, explained for NEET
When you dissolve a non-volatile solute in a solvent, the boiling point of the solution is higher than that of the pure solvent. This is boiling-point elevation — a colligative property that depends only on the number of solute particles, not their identity.
The trap that costs marks: an ionic compound like NaCl dissociates into ions. If a question hands you NaCl dissolved in water and you plug molality straight into ΔT_b = K_b · m, you get the answer for a non-electrolyte. The actual elevation is larger because NaCl produces two ions per formula unit — you must multiply by the Van 't Hoff factor i. Forgetting i is a common confusion in NEET colligative-property calculations.
The core relationship (NCERT Class 12 Chemistry Chapter 1, page 18):
ΔT_b = i · K_b · m
where ΔT_b is the elevation in boiling point, K_b is the ebullioscopic constant of the solvent (for water, K_b = 0.52 K·kg/mol), m is molality (moles of solute per kg of solvent), and i is the Van 't Hoff factor.
For non-electrolytes (glucose, urea), i = 1 and the formula reduces to ΔT_b = K_b · m. For electrolytes: NaCl → Na⁺ + Cl⁻ gives i ≈ 2; CaCl₂ → Ca²⁺ + 2Cl⁻ gives i ≈ 3.
Key points to lock in:
- Molality, not molarity — colligative formulas use moles of solute per kg of solvent, not per litre of solution. Molality is temperature-independent.
- K_b is a solvent property. Water's K_b = 0.52 K·kg/mol. Different solvents have different K_b values.
- Higher ΔT_b means more solute particles. Comparing 0.1 m NaCl (i = 2) vs 0.1 m glucose (i = 1): NaCl gives double the elevation.
- Boiling-point elevation can be used to determine molar mass of an unknown non-electrolyte solute: M₂ = (i · K_b · w₂ × 1000) / (ΔT_b · w₁), where w₂ is mass of solute and w₁ is mass of solvent in grams.
Watch out: when the question says "0.1 molal NaCl," the molality refers to the formula units dissolved, not the total ion concentration. The i factor handles the ion count separately.
Can you answer these Elevation Boiling 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 boiling-point elevation of a solution depends on:
Show answer and why every option is right or wrong
Answer: B. Boiling-point elevation is a colligative property — it depends on the number of solute particles, not their chemical identity (NCERT Class 12 Chemistry Chapter 1, page 18).
Why A is wrong: A is wrong because colligative properties by definition do not depend on the nature (identity) of the solute — only on the number of dissolved particles.
Why C is wrong: C is wrong because boiling-point elevation applies to non-volatile solutes that contribute negligible vapour pressure. The property depends on particle count, not solute vapour pressure.
Why D is wrong: D is wrong because while K_b is a solvent-specific constant, the elevation also depends on molality (solute particle count). 'Molecular mass of the solvent only' is incomplete and misleading.
MCQ 2Easy RecallPractice
The ebullioscopic constant K_b depends on:
Show answer and why every option is right or wrong
Answer: A. K_b is a property of the solvent alone — for water it is 0.52 K·kg/mol regardless of which solute is dissolved (NCERT Class 12 Chemistry Chapter 1, page 17).
Why B is wrong: B is wrong because K_b is determined entirely by the solvent's properties (its boiling point and enthalpy of vaporisation). Changing the solute does not change K_b.
Why C is wrong: C is wrong because K_b is a constant for a given solvent. It does not change with how concentrated the solution is.
Why D is wrong: D is wrong because K_b has nothing to do with the solute's molar mass. It is derived from the solvent's thermodynamic properties.
MCQ 3Easy RecallPractice
Which of the following is used as the unit of molality?
Show answer and why every option is right or wrong
Answer: C. Molality is defined as moles of solute per kilogram of solvent, so its unit is mol/kg (NCERT Class 11 Chemistry Chapter 1).
Why A is wrong: A is wrong because mol/L is the unit of molarity (moles per litre of solution), not molality.
Why B is wrong: B is wrong because g/L represents mass concentration, not molality.
Why D is wrong: D is wrong because mol/mol is dimensionless and corresponds to mole fraction, not molality.
MCQ 4Direct ApplicationPractice
3.0 g of urea (molar mass = 60 g/mol) is dissolved in 500 g of water. The boiling-point elevation is (K_b for water = 0.52 K·kg/mol):
Show answer and why every option is right or wrong
Answer: D. Moles of urea = 3.0/60 = 0.05 mol. Molality = 0.05/0.500 = 0.10 mol/kg. ΔT_b = K_b · m = 0.52 × 0.10 = 0.052 K. Urea is a non-electrolyte so i = 1.
Why A is wrong: A (5.2 K) results from using molality = 10 mol/kg — a gross unit error, likely dividing moles by grams without the kg conversion.
Why B is wrong: B (0.104 K) results from using mass of solvent as 250 g instead of 500 g, doubling the molality. Always read the solvent mass carefully.
Why C is wrong: C (0.52 K) results from using molality = 1.0 mol/kg — this happens if you forget to convert 500 g to 0.500 kg and instead divide 0.05 mol by 0.05 (mishandling units).
MCQ 5Direct ApplicationPractice
45 g of glucose (M = 180 g mol⁻¹) is dissolved in 500 g of water. Taking Kb for water as 0.52 K kg mol⁻¹, the elevation in boiling point is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Moles of glucose = 45/180 = 0.25 mol. Molality = moles per kilogram of solvent = 0.25/0.500 kg = 0.50 mol kg⁻¹. Glucose does not dissociate (i = 1), so ΔTb = Kb·m = 0.52 × 0.50 = 0.26 K.
Why B is wrong: B is wrong because 0.13 K uses the number of moles, 0.25, in place of the molality. Molality is per kilogram of solvent, and there is only half a kilogram here.
Why C is wrong: C is wrong because 2.6 × 10⁻⁴ K divides the moles by the solvent mass in grams, 500, instead of kilograms, 0.500.
Why D is wrong: D is wrong because 373.41 K is the new BOILING POINT (100.26 °C), not the elevation. The question asks for ΔTb, the rise itself.
MCQ 6Direct ApplicationPractice
0.1 molal aqueous solutions of NaCl and glucose are prepared. Which has a higher boiling point?
Show answer and why every option is right or wrong
Answer: C. NaCl dissociates (i ≈ 2), so its effective particle concentration is ~0.2 molal vs 0.1 molal for glucose (i = 1). Higher particle count → higher ΔT_b → higher boiling point for NaCl.
Why A is wrong: A is wrong because glucose is a non-electrolyte (i = 1) and produces fewer particles than NaCl at the same molality. Fewer particles means smaller ΔT_b (trap: forgetting the Van 't Hoff factor for the electrolyte).
Why B is wrong: B is wrong because NaCl is an electrolyte producing 2 ions per formula unit. Equal molality does not mean equal particle count when one solute dissociates.
Why D is wrong: D is wrong because K_b cancels when comparing two solutions in the same solvent. The comparison depends only on the effective molality (i × m), and NaCl has the higher value.
MCQ 7CalculationPractice
5.85 g of NaCl (molar mass = 58.5 g/mol) is dissolved in 250 g of water. Assuming complete dissociation, the boiling-point elevation is (K_b for water = 0.52 K·kg/mol):
Show answer and why every option is right or wrong
Answer: B. Moles of NaCl = 5.85/58.5 = 0.10 mol. Molality = 0.10/0.250 = 0.40 mol/kg. NaCl → Na⁺ + Cl⁻, so i = 2. ΔT_b = i · K_b · m = 2 × 0.52 × 0.40 = 0.416 K.
Why A is wrong: A (0.208 K) results from forgetting the Van 't Hoff factor — using ΔT_b = K_b · m = 0.52 × 0.40 = 0.208 K. NaCl is an electrolyte; you must multiply by i = 2 (trap: forgetting Van 't Hoff factor for ionic compounds).
Why C is wrong: C (0.832 K) results from using solvent mass as 125 g instead of 250 g, doubling the molality to 0.80 mol/kg, then correctly applying i = 2.
Why D is wrong: D (1.04 K) results from using molality = 1.0 mol/kg — likely dividing 0.10 mol by 0.10 kg instead of 0.250 kg, and then applying i = 2.
MCQ 8CalculationPractice
An aqueous solution of a non-electrolyte solute boils at 100.26°C. If K_b for water is 0.52 K·kg/mol, the molality of the solution is:
Show answer and why every option is right or wrong
Answer: D. ΔT_b = 100.26 − 100.00 = 0.26 K. For a non-electrolyte, i = 1. m = ΔT_b / K_b = 0.26 / 0.52 = 0.50 mol/kg.
Why A is wrong: A (0.13 mol/kg) is half of ΔT_b: it divides 0.26 by 2 as if applying i = 2 to a non-electrolyte, and never divides by K_b.
Why B is wrong: B (0.26 mol/kg) results from using K_b = 1.0 instead of 0.52, or equivalently from confusing ΔT_b with molality directly.
Why C is wrong: C (1.0 mol/kg) would raise the boiling point by 0.52 × 1.0 = 0.52 K, twice the observed 0.26 K.
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Elevation Boiling Point: quick recall before you leave
How do you solve a Elevation Boiling Point question? A worked example
- 1
Given
• Mass of urea (w₂) = 6.0 g• Molar mass of urea (M₂) = 60 g/mol (exact, problem-defined)• Mass of solvent (w₁) = 200 g = 0.200 kg• K_b for water = 0.52 K·kg/mol• Urea is a non-electrolyte → i = 1
- 2
Required
Boiling point of the solution (T_b).
- 3
Concept
Boiling-point elevation is a colligative property. A non-volatile solute raises the boiling point by ΔT_b = i · K_b · m. For non-electrolytes, i = 1.
- 4
Formula
ΔT_b = i · K_b · m, where m = n / (mass of solvent in kg)
- 5
Substitution
Moles of urea: n = 6.0 / 60 = 0.10 mol
Molality: m = 0.10 / 0.200 = 0.50 mol/kg
ΔT_b = 1 × 0.52 × 0.50 - 6
Calculation
ΔT_b = 0.52 × 0.50 = 0.26 K
Note: the molar mass 60 g/mol and the integer 1 (Van 't Hoff factor for a non-electrolyte) are exact values — they do not limit significant figures. The given data (6.0 g, 200 g, 0.52 K·kg/mol) are all 2 significant figures, so the answer is reported to 2 significant figures. - 7
Final answer
T_b = 100.00 + 0.26 = 100.26°C
The boiling point of the solution is 100.26°C (or equivalently, ΔT_b = 0.26 K). - 8
Common trap
If the solute were NaCl instead of urea at the same molality, forgetting to multiply by i = 2 would give exactly half the correct ΔT_b. For any ionic solute, always ask: "How many ions does one formula unit produce?"
- 9
Similar NEET-style question
"0.5 molal aqueous solution of CaCl₂ is prepared. Calculate the elevation in boiling point. Assume complete dissociation. (K_b = 0.52 K·kg/mol)."
Approach: CaCl₂ → Ca²⁺ + 2Cl⁻, so i = 3. ΔT_b = 3 × 0.52 × 0.5 = 0.78 K.
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What to remember before solving Elevation Boiling Point questions
Which Elevation Boiling Point formulas do you need for NEET?
Boiling-point elevation
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 |
Valid when
- Dilute solution
- Non-electrolyte
More in Solutions: 4 exam traps and mistakes · 7 formulas · 2 question patterns from its other lessons.
Elevation Boiling Point questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys.
NEET 2025
Which of the following aqueous solution will exhibit highest boiling point?
10.015M C6H12O6
20.01M Urea
30.01M KNO3
40.01M Na2SO4
NTA Answer: Option 4(final)
How does NEET ask about Elevation Boiling 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
NCERT refs: Class 12 Chemistry Chapter 1, p.18
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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