Kohlrausch Law

8 MCQs2 revision cards9-step worked example

Source: NCERT Redox ReactionsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity (Λ°_m) of an electrolyte equals the sum of the individual contributions of its cation and anion at infinite dilution (NCERT Class 12 Chemistry Chapter 2, page 20). Each ion contributes a fixed amount — λ°₊ for the cation, λ°₋ for the anion — regardless of the other ion present.

The formula: Λ°_m = ν₊ λ°₊ + ν₋ λ°₋

where ν₊ and ν₋ are the number of moles of cation and anion per formula unit. For NaCl: ν₊ = 1, ν₋ = 1. For BaCl₂: ν₊ = 1, ν₋ = 2.

Why NEET cares: You cannot measure Λ°_m of weak electrolytes directly — their conductivity never plateaus even at extreme dilution. Kohlrausch's law lets you calculate it from strong electrolyte data. This is the single most-tested application of this topic.

The standard trick for weak electrolytes:

To find Λ°_m(CH₃COOH): Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl)

The Na⁺ and Cl⁻ contributions cancel, leaving CH₃COO⁻ + H⁺.

Common distractors in NEET MCQs on this topic:

Forgetting the stoichiometric coefficient (ν). For CaCl₂, students write λ°(Ca²⁺) + λ°(Cl⁻) instead of λ°(Ca²⁺) + 2λ°(Cl⁻). The factor of 2 on chloride is load-bearing.
Inverting the combination. When computing Λ°_m of a weak electrolyte, students add the wrong pair of strong electrolytes and fail to cancel the auxiliary ions.
Unit confusion with the molar conductivity formula. When computing Λ_m = 1000κ/C, students either invert the C term or drop the 1000 factor. The pattern NEET pattern: molar conductivity problem confirms both as common NEET distractors.

Watch-out: Kohlrausch's law applies only at infinite dilution. At finite concentration, interionic interactions make ion contributions non-additive.

Practice 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

Kohlrausch's law states that at infinite dilution, each ion contributes to the total molar conductivity of an electrolyte:

MCQ 2Easy RecallPractice

The limiting molar conductivity of BaCl₂ at infinite dilution is expressed as:

MCQ 3Easy RecallPractice

The primary application of Kohlrausch's law in NEET-level problems is to:

MCQ 4Direct ApplicationPractice

Given: Λ°_m(NaCl) = 126.4 S cm² mol⁻¹, Λ°_m(NaNO₃) = 121.6 S cm² mol⁻¹, Λ°_m(KNO₃) = 145.0 S cm² mol⁻¹. The value of Λ°_m(KCl) in S cm² mol⁻¹ is:

MCQ 5Direct ApplicationPractice

Given: Λ°_m(CH₃COONa) = 91.0 S cm² mol⁻¹, Λ°_m(HCl) = 426.2 S cm² mol⁻¹, Λ°_m(NaCl) = 126.4 S cm² mol⁻¹. The limiting molar conductivity of acetic acid (CH₃COOH) in S cm² mol⁻¹ is:

MCQ 6Direct ApplicationPractice

The specific conductance (κ) of a 0.025 M solution of an electrolyte is 1.25 × 10⁻⁴ S cm⁻¹. Its molar conductivity (Λ_m) in S cm² mol⁻¹ is:

MCQ 7Concept TrapPractice

Why can the limiting molar conductivity of acetic acid not be determined by direct measurement of conductivity at progressively lower concentrations?

MCQ 8CalculationPractice

Given: Λ°_m(Ba(OH)₂) = 457.6 S cm² mol⁻¹, Λ°_m(BaCl₂) = 280.0 S cm² mol⁻¹, Λ°_m(NH₄Cl) = 149.8 S cm² mol⁻¹. Using Kohlrausch's law, the value of Λ°_m(NH₄OH) in S cm² mol⁻¹ is:

Quick recall before you leave

Worked Example

Pattern: NEET pattern: molar conductivity problem — Compute Λ°_m of a weak electrolyte from strong electrolyte data via Kohlrausch's law.

1
Given
Λ°_m(CH₃COONa) = 91.0 S cm² mol⁻¹ Λ°_m(HCl) = 426.2 S cm² mol⁻¹ Λ°_m(NaCl) = 126.4 S cm² mol⁻¹ Find: Λ°_m(CH₃COOH)
2
Required
Limiting molar conductivity of acetic acid, a weak electrolyte whose Λ°_m cannot be measured directly.
3
Concept
Kohlrausch's law: Λ°_m = ν₊λ°₊ + ν₋λ°₋. Each ion's contribution is independent. By choosing strong electrolytes that share the target ions, auxiliary ions cancel in the sum.
4
Formula
Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl) **Why this works:** - CH₃COONa contributes λ°(CH₃COO⁻) + λ°(Na⁺) - HCl contributes λ°(H⁺) + λ°(Cl⁻) - NaCl contributes λ°(Na⁺) + λ°(Cl⁻) Subtracting NaCl cancels λ°(Na⁺) and λ°(Cl⁻), leaving λ°(CH₃COO⁻) + λ°(H⁺) = Λ°_m(CH₃COOH).
5
Substitution
Λ°_m(CH₃COOH) = 91.0 + 426.2 − 126.4
6
Calculation
= 517.2 − 126.4 = 390.8 S cm² mol⁻¹ All given values are exact data from conductivity tables; no sig-fig ambiguity arises.
7
Final answer
**Λ°_m(CH₃COOH) = 390.8 S cm² mol⁻¹**
8
Common trap
**Forgetting the subtraction step.** If you add all three values (91.0 + 426.2 + 126.4 = 643.6), you get a distractor that double-counts Na⁺ and Cl⁻. The cancellation method requires one subtraction. If you forget it, auxiliary ions remain and the answer is inflated by ~65%. For electrolytes with ν > 1 (e.g., Ba(OH)₂), remember to apply the stoichiometric coefficient when extracting individual λ° values.
9
Similar NEET-style question
Given: Λ°_m(KCl) = 149.8 S cm² mol⁻¹, Λ°_m(KNO₃) = 145.0 S cm² mol⁻¹, Λ°_m(AgNO₃) = 133.4 S cm² mol⁻¹. Find Λ°_m(AgCl). [Answer: 149.8 + 133.4 − 145.0 = 138.2 S cm² mol⁻¹] ---

Before solving, remember these

Law

Kohlrausch's law of independent migration

Λ°_m (molar conductivity at infinite dilution) = ν₊λ°₊ + ν₋λ°₋. Allows determination of Λ°_m for weak electrolytes from strong electrolytes.

-- NCERT Class 12 Chemistry, Ch. 2, p. 20

Formulas

Cell EMF

E_{cell}^{\circ} = E_{cath}^{\circ} - E_{anode}^{\circ}

Both as reduction potentials. E°_cell > 0 → spontaneous.

Symbol	Quantity	SI Unit
E°_cell	standard cell EMF	V
E°_red	reduction potential	V

Valid when

Standard conditions (1 M, 1 bar, 298 K)
Both half-reactions as reductions

Faraday's law of electrolysis

m = \frac{M I t}{n F}

Mass deposited at electrode. M = molar mass; I = current; t = time; n = electrons per ion.

Symbol	Quantity	SI Unit
m	mass deposited	g
M	molar mass	g/mol
I	current	A
t	time	s
n	electrons per ion	-
F	Faraday	C/mol

Valid when

Steady current
Single product

ΔG from EMF

Δ G = - n F E

Connection between thermodynamics and electrochemistry. F = 96485 C/mol.

Symbol	Quantity	SI Unit
ΔG	Gibbs energy change	J
n	electrons transferred	-
F	Faraday 96485	C/mol
E	cell EMF	V

Valid when

Single redox process

Molar conductivity

Λ_{m} = \frac{1000 κ}{C}

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

Nernst equation

E = E^{\circ} - \frac{0.0591}{n} lo g Q

Cell potential at non-standard conditions. n = electrons transferred. At equilibrium E=0, Q=K.

Symbol	Quantity	SI Unit
E	cell potential	V
E°	standard	V
n	electrons	-
Q	reaction quotient	-

Valid when

298 K (else use RT/F)
Single redox process

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Nernst n electrons cumulative error

Category: Negative Marking

Multi-step Nernst problem: identify electrons, write Q correctly, plug into 0.0591/n. Each sub-step has factor errors.

When it triggers

Cell EMF problem at non-standard conditions.

How to avoid

Step-by-step: (1) write balanced redox; (2) count n electrons; (3) compute Q from concentrations; (4) plug into Nernst. Verify by checking limits: at standard conditions Q=1, log Q=0, E=E°.

Trap

KMnO4 oxidation product depends on medium

Category: Inorganic Exception

Student assumes Mn²⁺ is the product regardless of medium. Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).

When it triggers

Question gives KMnO4 oxidation in unspecified or specific medium.

How to avoid

Always check medium. In acidic: Mn(+7) → Mn(+2). In neutral: → Mn(+4) (MnO₂). In basic: → Mn(+6) (manganate). The number of electrons (n) in Nernst calculations depends accordingly.

Mistake

Uses n=1 in Faraday's law for ions like Cu²⁺ (actual n=2) or Al³⁺ (n=3).

Root cause: concept gap

Correction

n = number of electrons per ion to deposit. Cu²⁺ + 2e⁻ → Cu: n=2. Al³⁺ + 3e⁻: n=3. m = MIt/(nF).

Mistake

Uses incorrect electron count n in Nernst equation.

Root cause: formula misuse

Correction

n = electrons transferred per balanced redox equation. For Cu²⁺ + 2e⁻ → Cu: n=2. For Mn(VII) → Mn(II): n=5.

Past Year Questions

17 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

If the rate constant of a reaction is 0.03 s–1, how much time does it take for 7.2 mol L–1 concentration of the reactant to get reduced to 0.9 mol L–1? (Given: log 2 = 0.301)

121.0 s

269.3 s

323.1 s

4210 s

NTA Answer: Option 2(final)

NEET 2025

Consider the following compounds : KO , H O and H SO 2 2 2 2 4 The oxidation state of the underlined elements in them are, respectively,

1+4, –4, and +6

2+1, –1, and +6

3+2, –2, and +6

4+1, –2, and +4

NTA Answer: Option 2(final)

NEET 2025

Match List-I with List-II List-I List-II (Example) (Type of Solution) A. Humidity I. Solid in solid B. Alloys II. Liquid in gas C. Amalgams III. Solid in gas D. Smoke IV. Liquid in solid Choose the correct answer from the options given below:

1A-III, B-II, C-I, D-IV

2A-II, B-IV, C-I,D-III

3A-II, B-I, C-IV, D-III

4A-III, B-I, C-IV, D-II

NTA Answer: Option 3(final)

NEET 2024Revised key

Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:

1Both Statement I and Statement II are correct.

2Both Statement I and Statement II are incorrect.

3Statement I is correct but Statement II is incorrect.

4Statement I is incorrect but Statement II is correct.

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

Match List I with List II. List I List II (Conversion) (Number of Faraday required) A. 1 mol of H O to O I. 3F 2 2 B. 1 mol of MnO− 4 to Mn2+ II. 2F C. 1.5 mol of Ca from molten CaCl III. 1F 2 D. 1 mol of FeO to Fe O IV. 5F 2 3 Choose the correct answer from the options given below:

1A-II, B-IV, C-I, D-III

2A-III, B-IV, C-I, D-II

3A-II, B-III, C-I, D-IV

4A-III, B-IV, C-II, D-I

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol–1, 1 F = 96487 C)

13.15 g

20.315 g

331.5 g

40.0315 g

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

The products A and B obtained in the following reactions, respectively, are 3ROH + PCl → RCl + A 3 ROH + PCl → RCl + HCl + B 5

1POCl and H PO 3 3 3

2POCl and H PO 3 3 4

3H PO and POCl 3 4 3

4H PO and POCl 3 3 3

NTA Answer: Option 4(revised_final)

NEET 2023

The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm–1 cm–1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of cell constant is

13.28 cm–1

21.26 cm–1

33.34 cm–1

41.34 cm–1

NTA Answer: Option 2(final)

NEET 2023

The number of  bonds,  bonds and lone pair of electrons in pyridine, respectively are:

112, 3, 0

211, 3, 1

312, 2, 1

411, 2, 0

NTA Answer: Option 2(final)

NEET 2022

At 298 K, the standard electrode potentials of Cu2+ / Cu, Zn2+ / Zn, Fe2+ / Fe and Ag+ / Ag are 0.34 V, –0.76 V, –0.44 V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?

12CuSO (aq) + 2Ag(s) → 2Cu(s) + Ag SO (aq) 4 2 4

2CuSO (aq) + Zn(s) → ZnSO (aq) + Cu(s) 4 4

3CuSO (aq) + Fe(s) → FeSO (aq) + Cu(s) 4 4

4FeSO (aq) + Zn(s) → ZnSO (aq) + Fe(s) 4 4

NTA Answer: Option 1(final)

NEET 2022

Which of the following statement is not correct about diborane?

1Both the Boron atoms are sp2 hybridised.

2There are two 3-centre-2-electron bonds.

3The four terminal B-H bonds are two centre two electron bonds.

4The four terminal Hydrogen atoms and the two Boron atoms lie in one plane.

NTA Answer: Option 1(final)

NEET 2022

Given below are half cell reactions: - 13 - M n O −4 + 8 H + + 5 e − → M n 2 + + 4 H 2 O E ºM n 2 + /M n O −4 = − 1 .5 1 0 V 1 2 O 2 + 2 H + + 2 e − → H 2 O Eº =+1.223V O2/H2O Will the permanganate ion, M n O –4 liberate O from water in the presence of an acid? 2

1No, because E cell =−2.733V

2Yes, because E c e ll = + 0 .2 8 7 V

3No, because E c e ll = – 0 .2 8 7 V

4Yes, because E c e ll = + 2 .7 3 3 V

NTA Answer: Option 2(final)

NEET 2022

If radius of second Bohr orbit of the He+ ion is 105.8 pm, what is the radius of third Bohr orbit of Li2+ ion?

1158.7 Å

2158.7 pm

315.87 pm

41.587 pm

NTA Answer: Option 2(final)

NEET 2021

Which one among the following is the correct option for right relationship between C and C for one mole P V of ideal gas?

1C = RC V P

2C + C = R P V

3C – C = R P V

4C = RC P V

NTA Answer: Option 3(final)

NEET 2021

The molar conductance of NaCl, HCl and CH COONa 3 at infinite dilution are 126.45, 426.16 and cm2 mol–1 91.0 S respectively. The molar conductance of CH COOH at infinite dilution is. 3 Choose the right option for your answer. cm2 mol–1

1540.48 S cm2 mol–1

2201.28 S cm2 mol–1

3390.71 S cm2 mol–1

4698.28 S

NTA Answer: Option 3(final)

NEET 2021

Which one of the following polymers is prepared by addition polymerisation?

1Dacron

2Teflon

3Nylon-66

4Novolac

NTA Answer: Option 2(final)

NEET 2021

The molar conductivity of 0.007 M acetic acid is 20 cm2 mol–1. S What is the dissociation constant of acetic acid? Choose the correct option. Λ° =350Scm2 mol−1  H+   Λ° mol−1 =50Scm2  CHCOO−  3 2.50×10−5 molL−1

11.75×10−4 molL−1

22.50×10−4 molL−1

31.75×10−5 molL−1

4Answer

NTA Answer: Option 4(final)

How NEET usually asks this

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

Apply Nernst equation to compute non-standard cell EMF given Q. n = electrons transferred.

Multi StepMedium

Common distractors

wrong n electrons

Uses incorrect electron count from half-reactions

Sources

NCERT refs: Class 12 Chemistry Chapter 2, p.20

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