Q.79 Chemistry ⭐ 360° Option Elimination Active
NEET 2025 Chemistry Question 79 Solution
Topic: Solutions · Official NTA Booklet Code 45
Question
Which of the following aqueous solution will exhibit highest boiling point?
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Verified Against NTA Final Answer Key
Option (3) : 0.01M Na₂SO₄
📚 Solutions Easy
📋 Given Parameters & Problem Setup
Solution (1) 0.01 M Urea (non-electrolyte, i = 1)
Solution (2) 0.01 M KNO₃ (binary electrolyte, i = 2)
Solution (3) 0.01 M Na₂SO₄ (ternary electrolyte, i = 3)
Solution (4) 0.015 M C₆H₁₂O₆ (Glucose, non-electrolyte, i = 1)
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⏱️ Save ~30 seconds The 30-Second Exam Hall Shortcut: ⚡ i × M Maximizer Shortcut
Boiling point elevation ΔT_b ∝ (i × M)! Urea = 1 × 0.01 = 0.01. KNO₃ = 2 × 0.01 = 0.02. Na₂SO₄ = 3 × 0.01 = 0.03. Glucose = 1 × 0.015 = 0.015! Na₂SO₄ has the LARGEST i × M (0.03) ⟹ HIGHEST BOILING POINT!
1 Elevation in boiling point formula: ΔT_b = i × K_b × m ≈ i × K_b × M for dilute aqueous solutions.
2 Higher ΔT_b means higher boiling point: T_b = T_b° + ΔT_b.
3 Calculate effective particle concentration (i × M) for each candidate:
4 Solution (1) Urea: i = 1 ⟹ i × M = 1 × 0.01 = 0.01 M.
5 Solution (2) KNO₃: Dissociates into K⁺ + NO₃⁻ (2 ions) ⟹ i = 2 ⟹ i × M = 2 × 0.01 = 0.02 M.
6 Solution (3) Na₂SO₄: Dissociates into 2 Na⁺ + SO₄²⁻ (3 ions) ⟹ i = 3 ⟹ i × M = 3 × 0.01 = 0.03 M.
7 Solution (4) Glucose (C₆H₁₂O₆): i = 1 ⟹ i × M = 1 × 0.015 = 0.015 M.
8 Compare values: 0.03 M (Na₂SO₄) > 0.02 M (KNO₃) > 0.015 M (Glucose) > 0.01 M (Urea).
9 Select Option (3) in under 10 seconds!
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Why You Can 100% Trust This Shortcut: Colligative Properties & van ’t Hoff Factor (NCERT Chemistry · Solutions: ΔT_b = i K_b m).
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Step-by-Step Rigorous Workflow & Mathematical Proof
Step 1
Colligative Principle Governing Boiling Point Elevation
The elevation in boiling point ΔT_b of an electrolyte solution depends on the total concentration of dissolved solute particles:
- • For dilute aqueous solutions, molality m is approximately equal to molarity M.
- • The solution exhibiting the greatest elevation ΔT_b will have the highest boiling point T_b.
Step 2
Determine van ’t Hoff Factor (i) and Effective Concentration for Each Solute
Assuming complete dissociation for strong electrolytes:
- • (1) Urea: Non-electrolyte ⟹ i = 1 ⟹ i × M = 0.01 M.
- • (2) KNO₃: Dissociates into K⁺ + NO₃⁻ ⟹ i = 2 ⟹ i × M = 2 × 0.01 = 0.02 M.
- • (3) Na₂SO₄: Dissociates into 2 Na⁺ + SO₄²⁻ ⟹ i = 3 ⟹ i × M = 3 × 0.01 = 0.03 M.
- • (4) Glucose: Non-electrolyte ⟹ i = 1 ⟹ i × M = 1 × 0.015 = 0.015 M.
Step 3
Rank Boiling Points and Select Matching Option
Rank solutions by decreasing effective particle concentration:
- • 0.01 M Na₂SO₄ produces the greatest particle concentration and thus the highest boiling point.
- • Matches Option (3) with complete precision.
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4-Option Deep Scan
360° Option Elimination Engine Key Feature
Diagnose every choice: Know exactly why the right option works and why distractors fail.
1 A
✗ Incorrect Verdict & Scientific Reason:
Incorrect. Produces the lowest boiling point (lowest i × M = 0.01 M).
⚠️
Student Fallacy (Why candidates pick this trap): Ignoring dissociation of ionic salts.
2 B
✗ Incorrect Verdict & Scientific Reason:
Incorrect. Produces only 2 ions per formula unit (i × M = 0.02 M).
⚠️
Student Fallacy (Why candidates pick this trap): Failing to realize Na₂SO₄ produces 3 ions.
3 C
✓ 100% Correct Verdict & Scientific Reason:
100% correct. Dissociates into 3 ions, yielding the maximum effective particle concentration (0.03 M).
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Why this is airtight: None; textbook application of colligative laws.
4 D
✗ Incorrect Verdict & Scientific Reason:
Incorrect. Although molarity is 0.015 M, it does not ionize (i = 1), so 0.015 < 0.03 M.
⚠️
Student Fallacy (Why candidates pick this trap): Looking only at nominal molarity (0.015 > 0.01) without multiplying by i.
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NTA Trap Alert: Nominal Concentration vs Effective Concentration (Glucose Trap)
Students notice that glucose has the highest numerical concentration (0.015 M vs 0.01 M) and hastily pick Option (4). They forget to multiply by van ’t Hoff factor i! Na₂SO₄ gives 3 × 0.01 = 0.03 M, which is DOUBLE the glucose concentration!
🛡️ How to avoid the -1 mark: ALWAYS multiply molarity by van ’t Hoff factor: Effective Molality = i × M!
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Core NEET Syllabus Standard
NEET Chemistry Core Syllabus · Solutions Standard Benchmark
Core Scientific Principle Tested:
"ΔT_b = i K_b m; highest i × M gives highest boiling point."
Syllabus Focus: Colligative Properties & Abnormal Molar Mass (van ’t Hoff Factor)
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
Multiply by ions: Na₂SO₄ has 3 ions ⟹ 3 × 0.01 = 0.03 M (Highest boiling point, Option 3)!
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