Definition
Molar conductivity
Λ_m = κ × 1000/C, where κ is specific conductance and C is molar concentration. Units: S·cm²·mol⁻¹. Λ_m increases with dilution (more ions free to move).
-- NCERT Class 12 Chemistry, Ch. 2, p. 45Molar Conductivity
8 MCQs4 revision cards9-step worked example
Source: NCERT Redox Reactions and ElectrochemistryPYQ coverage: NEET 2021Official key: NTA-verifiedLast updated: 25 Sep 2026
Molar Conductivity, explained for NEET
The high-frequency trap in molar conductivity problems is the unit-conversion factor of 1000. Students who write Λ_m = κ × C (instead of κ × 1000 / C) lose marks on what should be a straightforward substitution.
Molar conductivity (Λ_m) is defined as the conductance of all ions produced by one mole of electrolyte when placed between electrodes 1 cm apart (NCERT Class 12 Chemistry Chapter 2, page 47). The operational formula is:
Λ_m = κ × 1000 / C
where κ is specific conductance (S cm⁻¹) and C is molar concentration (mol L⁻¹). The factor 1000 converts litres to cm³ (1 L = 1000 cm³). The SI unit of Λ_m is S cm² mol⁻¹.
Variation with concentration:
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Strong electrolytes (NaCl, KCl, HCl): Λ_m increases slightly with dilution because inter-ionic attraction decreases. Even at moderate concentrations, dissociation is essentially complete. Λ_m approaches Λ°_m (limiting molar conductivity) linearly per the Debye-Hückel-Onsager equation.
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Weak electrolytes (CH₃COOH, NH₄OH): Λ_m increases steeply with dilution because the degree of dissociation (α) itself increases. At infinite dilution α → 1, but Λ°_m cannot be obtained by extrapolation — it must be calculated using Kohlrausch's law.
Watch-out: When κ is given in S m⁻¹ (not S cm⁻¹), the conversion factor changes. Always check units before substituting. The pattern "inverts C multiplier" (writing κC instead of κ/C) accounts for a common distractor in NEET papers.
Can you answer these Molar Conductivity MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
The specific conductance of a 0.02 M KCl solution is 2.768 × 10⁻³ S cm⁻¹. What is the molar conductivity?
Show answer and why every option is right or wrong
Answer: D. Λ_m = κ × 1000 / C = 2.768 × 10⁻³ × 1000 / 0.02 = 138.4 S cm² mol⁻¹ (NCERT Class 12 Chemistry Chapter 2, page 47).
Why A is wrong: This drops one factor of 10 in the calculation, giving 138.4/10 = 13.84.
Why B is wrong: This results from computing κ × C directly without the 1000 factor — the 'inverts-C-multiplier' distractor.
Why C is wrong: This omits the factor of 1000, giving κ/C = 2.768 × 10⁻³ / 0.02 = 0.1384, then misplacing the decimal.
MCQ 2Concept TrapPractice
On diluting a weak electrolyte solution from 0.1 M to 0.001 M, what happens to molar conductivity?
Show answer and why every option is right or wrong
Answer: D. For weak electrolytes, dilution increases the degree of dissociation α, which sharply raises Λ_m. The rise is steep, not linear (NCERT Class 12 Chemistry Chapter 2, page 47).
Why A is wrong: Molar conductivity never decreases with dilution for electrolytes — dilution cannot reduce ion mobility or dissociation.
Why B is wrong: Λ_m of a weak electrolyte is NOT concentration-independent; it varies strongly with α.
Why C is wrong: Linear increase applies to strong electrolytes (Debye-Hückel-Onsager equation). Weak electrolytes show a steep, non-linear rise.
MCQ 3Easy RecallPractice
If conductivity κ is expressed in S cm⁻¹ and concentration in mol cm⁻³, the unit of molar conductivity (Λ_m) is:
Show answer and why every option is right or wrong
Answer: B. Λ_m = κ × volume per mole. Units: (S cm⁻¹)(cm³ mol⁻¹) = S cm² mol⁻¹ (NCERT Class 12 Chemistry Chapter 2, page 47).
Why A is wrong: S cm⁻¹ is the unit of specific conductance (κ), not molar conductivity.
Why C is wrong: Ω (ohm) is resistance; conductivity uses S (siemens). Ω cm² mol⁻¹ inverts the conductance unit.
Why D is wrong: The correct dimensional combination places cm² in the numerator, not the denominator.
MCQ 4Easy RecallPractice
For a strong electrolyte, which statement about the Λ_m vs √C plot is correct?
Show answer and why every option is right or wrong
Answer: A. The Debye-Hückel-Onsager equation gives Λ_m = Λ°_m − A√C for strong electrolytes — a linear decrease with √C (NCERT Class 12 Chemistry Chapter 2, page 47).
Why B is wrong: Λ_m of strong electrolytes does depend on concentration (through inter-ionic forces); only at infinite dilution does it reach the constant Λ°_m.
Why C is wrong: Increasing inter-ionic effects at higher concentration reduce Λ_m, not increase it.
Why D is wrong: The relationship is linear (Λ°_m − A√C), not exponential. Exponential decay has no theoretical basis here.
MCQ 5Direct ApplicationPractice
If the specific conductance of a 0.5 M solution is 1.0 × 10⁻² S cm⁻¹, and of a 0.05 M solution is 2.0 × 10⁻³ S cm⁻¹, which solution has higher Λ_m?
Show answer and why every option is right or wrong
Answer: C. For 0.5 M: Λ_m = 1.0 × 10⁻² × 1000 / 0.5 = 20. For 0.05 M: Λ_m = 2.0 × 10⁻³ × 1000 / 0.05 = 40. The dilute solution has higher Λ_m (NCERT Class 12 Chemistry Chapter 2, page 47).
Why A is wrong: This gives the correct value for 0.5 M (20) but incorrectly claims it is higher. The 0.05 M solution yields Λ_m = 40, which is larger.
Why B is wrong: The two values are 20 and 40 — clearly unequal. Equal Λ_m would require κ/C to be the same ratio in both cases.
Why D is wrong: The formula Λ_m = κ × 1000/C is sufficient. No additional information about electrolyte type is needed for numerical comparison.
MCQ 6Concept TrapPractice
Why can't Λ°_m of a weak electrolyte like acetic acid be determined by extrapolating the Λ_m vs √C curve?
Show answer and why every option is right or wrong
Answer: A. The Λ_m vs √C curve for weak electrolytes is steep and non-linear near low C (due to rapidly increasing α), so linear extrapolation gives unreliable values. Kohlrausch's law is used instead (NCERT Class 12 Chemistry Chapter 2, page 47).
Why B is wrong: Λ_m of weak electrolytes varies significantly with concentration — this is the very reason extrapolation is attempted.
Why C is wrong: At infinite dilution, a weak electrolyte is fully dissociated (α = 1) and does conduct. The issue is experimental access to that limit, not absence of conductance.
Why D is wrong: Λ°_m is well-defined for weak electrolytes — it represents the molar conductivity at complete dissociation. The issue is obtaining it experimentally by extrapolation.
MCQ 7Direct ApplicationPractice
A student calculates Λ_m using the formula Λ_m = κ × C. For a 0.1 M NaCl solution with κ = 1.07 × 10⁻² S cm⁻¹, what answer does this incorrect formula give, and what is the correct answer?
Show answer and why every option is right or wrong
Answer: C. Incorrect formula κ × C = 1.07 × 10⁻² × 0.1 = 1.07 × 10⁻³. Correct: κ × 1000/C = 1.07 × 10⁻² × 1000/0.1 = 107 S cm² mol⁻¹. The factor error is 10⁵ (NCERT Class 12 Chemistry Chapter 2, page 47).
Why A is wrong: This swaps the correct and incorrect answers. κ × C gives 1.07 × 10⁻³ (too small), not 107.
Why B is wrong: The correct answer is 107, not 1.07 × 10⁻¹. The factor 1000/C = 1000/0.1 = 10000, applied to κ gives 107.
Why D is wrong: The incorrect formula κ × C yields 1.07 × 10⁻³, not 10.7. The student must multiply by C, not by 1000×C to get the wrong answer.
MCQ 8CalculationPractice
Given κ = 3.5 × 10⁻³ S cm⁻¹ for a 0.025 M solution, calculate Λ_m. If the concentration is halved (0.0125 M) and κ drops to 2.0 × 10⁻³ S cm⁻¹, what is the new Λ_m?
Show answer and why every option is right or wrong
Answer: B. Initial: 3.5 × 10⁻³ × 1000/0.025 = 140. New: 2.0 × 10⁻³ × 1000/0.0125 = 160 S cm² mol⁻¹. Λ_m increases on dilution as expected (NCERT Class 12 Chemistry Chapter 2, page 47).
Why A is wrong: The second value (80) comes from dividing by the OLD concentration: 2.0 × 10⁻³ × 1000/0.025 = 80. The new Λ_m must use the new concentration, 0.0125 M.
Why C is wrong: This omits the 1000 factor in both calculations: 3.5 × 10⁻³/0.025 = 0.14 and 2.0 × 10⁻³/0.0125 = 0.16. The 1000 multiplier is mandatory for L-to-cm³ conversion.
Why D is wrong: The initial Λ_m = 3.5 × 10⁻³ × 1000/0.025 = 140, not 87.5. Getting 87.5 implies dividing by 0.04 instead of 0.025.
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Molar Conductivity: quick recall before you leave
How do you solve a Molar Conductivity question? A worked example
Pattern: NEET pattern: molar conductivity problem — compute Λ_m from κ and C.
- 1
Given
The specific conductance of a 0.040 M KCl solution is κ = 5.456 × 10⁻³ S cm⁻¹.
- 2
Required
Calculate the molar conductivity Λ_m in S cm² mol⁻¹.
- 3
Concept
Molar conductivity relates specific conductance to the amount of electrolyte. The 1000 factor accounts for volume conversion from litres to cm³.
- 4
Formula
Λ_m = κ × 1000 / C
- 5
Substitution
Λ_m = (5.456 × 10⁻³ S cm⁻¹) × 1000 (cm³ L⁻¹) / 0.040 (mol L⁻¹)
- 6
Calculation
Numerator: 5.456 × 10⁻³ × 1000 = 5.456, with units S cm⁻¹ × cm³ L⁻¹. Track the units deliberately here, because this is where the factor of 1000 earns its place: the 1000 converts litres to cm³, so S cm⁻¹ × cm³ mol⁻¹ leaves S cm² mol⁻¹ once the concentration in mol L⁻¹ is divided out.
κ has units S cm⁻¹. Multiply by 1000 cm³ L⁻¹ gives 5.456 S cm² L⁻¹. Divide by 0.040 mol L⁻¹:
Λ_m = 5.456 / 0.040 = 136.4 S cm² mol⁻¹
Note on exact values: The concentration 0.040 M and the factor 1000 are defined exact values (problem statement and unit conversion respectively). They do not limit significant figures. The answer precision is governed by κ (4 significant figures). - 7
Final answer
Λ_m = 136.4 S cm² mol⁻¹
- 8
Common trap
The "inverts-C-multiplier" distractor: writing Λ_m = κ × C = 5.456 × 10⁻³ × 0.040 = 2.18 × 10⁻⁴ — off by a factor of ~6.25 × 10⁵ from the correct answer. Another common error is forgetting the 1000 factor, which gives 5.456 × 10⁻³ / 0.040 = 0.1364 (off by ×1000).
- 9
Similar NEET-style question
"The specific conductance of 0.1 M CH₃COOH is 5.0 × 10⁻⁴ S cm⁻¹. Calculate Λ_m and, given Λ°_m = 390.5 S cm² mol⁻¹, determine the degree of dissociation α."
(Answer: Λ_m = 5.0; α = Λ_m / Λ°_m = 5.0 / 390.5 ≈ 0.0128)
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What to remember before solving Molar Conductivity questions
Which Molar Conductivity formulas do you need for NEET?
Molar conductivity
Molar conductivity from specific conductance. Increases with dilution as more ions are free.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Λ_m | molar conductivity | S cm^2/mol |
| κ | specific conductance | S/cm |
| C | molarity | mol/L |
Valid when
- Aqueous solution
- Single electrolyte
More in Redox Reactions and Electrochemistry: 4 exam traps and mistakes · 4 formulas · 1 question pattern from its other lessons.
Molar Conductivity questions from past NEET papers
1 question from NEET 2021. Answers verified against NTA official keys.
NEET 2021
12.50 × 10⁻⁵ mol L⁻¹
21.75 × 10⁻⁴ mol L⁻¹
32.50 × 10⁻⁴ mol L⁻¹
41.75 × 10⁻⁵ mol L⁻¹
NTA Answer: Option 4(final)
All 21 past-paper questions from Redox Reactions and Electrochemistry →
How does NEET ask about Molar Conductivity?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Compute Λ_m from κ and C. Or compute Λ°_m of weak electrolyte from strong electrolyte values via Kohlrausch.
Direct ApplicationMedium
Common distractors
inverts c multiplier
Uses Λ = κC instead of κ/C
forgets 1000 factor
Drops 1000 in unit conversion
Sources
NCERT refs: Class 12 Chemistry Chapter 2, p.47
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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