ChemistryCBSE Class 12 · Electrochemistry

Conductivity, Molar Conductivity and Kohlrausch's Law

Updated 15 October 2026 · Fact-checked

Conductivity ($\kappa$) is the conductance of 1 cm³ of solution; molar conductivity is $\Lambda_m = \frac{\kappa \times 1000}{c}$ in S cm² mol⁻¹. On dilution $\kappa$ falls but $\Lambda_m$ rises. Kohlrausch's law says limiting molar conductivity is the sum of the ions' contributions, which gives $\Lambda^\circ_m$ and the degree of dissociation of weak electrolytes.

Understand Conductivity, Molar Conductivity and Kohlrausch's Law

An electrolyte solution conducts because its ions move. To compare solutions we need quantities that do not depend on the size of the container.

Resistance to conductivity. Resistance $R$ (ohm, Ω) of a column of solution is $R = ρ \frac{l}{A}$ , where $ρ$ is resistivity. The inverse quantities are conductance $G = 1/ R$ (siemens, S) and conductivity $κ = 1/ ρ$ (S cm⁻¹ or S m⁻¹). So $κ = \frac{1}{R} \cdot \frac{l}{A}$ . In a conductivity cell, $l / A$ is fixed by the cell's construction and is called the cell constant $G^{*}$ ; thus $κ = \frac{G ^{*}}{R}$ . The cell constant is found by first measuring the resistance of a KCl solution whose conductivity is accurately known. Measurement uses an alternating current source (a Wheatstone bridge arrangement), because direct current would electrolyse the solution and change it.

Molar conductivity. Conductivity depends on how many ions are in the volume, so it is unfair for comparing different concentrations. Molar conductivity $Λ_{m} = \frac{κ}{c}$ gives the conducting power of all the ions from one mole of electrolyte. With $κ$ in S cm⁻¹ and $c$ in mol L⁻¹, $Λ_{m} = \frac{κ \times 1000}{c}$ S cm² mol⁻¹.

Effect of dilution. On dilution, conductivity decreases because the number of ions per cm³ falls. Molar conductivity increases, because $Λ_{m}$ now refers to a larger volume containing one mole, and the ions move more freely or more of the electrolyte ionises. When $c$ approaches zero, $Λ_{m}$ approaches the limiting molar conductivity $Λ_{m}^{\circ}$ .

Strong vs weak electrolytes. For strong electrolytes such as KCl, $Λ_{m}$ rises slowly and linearly with $c$ : $Λ_{m} = Λ_{m}^{\circ} - A c$ (the Debye–Hückel–Onsager form). Extrapolating the line to $c = 0$ gives $Λ_{m}^{\circ}$ . The slope $A$ depends on the charges of the ions: it is the same for all 1-1 electrolytes such as NaCl and KBr. For weak electrolytes such as ethanoic acid, $Λ_{m}$ is low at ordinary concentrations and rises very steeply near infinite dilution, because the degree of dissociation increases sharply. The curve cannot be extrapolated, so $Λ_{m}^{\circ}$ must be obtained another way.

Kohlrausch's law of independent migration of ions. At infinite dilution every ion moves independently, so the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each multiplied by the number of such ions: $Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}$ . For example $Λ_{m}^{\circ} (C a C l_{2}) = λ^{\circ} (C a^{2 +}) + 2 λ^{\circ} (C l^{-})$ .

Applications. (1) $Λ_{m}^{\circ}$ of a weak electrolyte by combining strong electrolytes: $Λ_{m}^{\circ} (C H_{3} C O O H) = Λ_{m}^{\circ} (H C l) + Λ_{m}^{\circ} (C H_{3} C O O N a) - Λ_{m}^{\circ} (N a C l)$ . (2) The degree of dissociation, $α = \frac{Λ _{m}}{Λ _{m}^{\circ}}$ , and from it the dissociation constant $K_{a} = \frac{c α ^{2}}{1 - α}$ .

Conductivity, Molar Conductivity and Kohlrausch's Law: formulas and key facts

Conductivity from resistance: $κ = \frac{1}{R} \cdot \frac{l}{A} = \frac{G ^{*}}{R}$; Cell constant $G^{*} = l / A$ in cm⁻¹; $κ$ in S cm⁻¹.
Molar conductivity: $Λ_{m} = \frac{κ \times 1000}{c}$; $κ$ in S cm⁻¹, $c$ in mol L⁻¹, answer in S cm² mol⁻¹. In SI: $Λ_{m} = κ / c$ with $c$ in mol m⁻³ gives S m² mol⁻¹.
Strong electrolytes (Debye–Hückel–Onsager): $Λ_{m} = Λ_{m}^{\circ} - A c$; Straight line against $c$ ; intercept $Λ_{m}^{\circ}$ .
Kohlrausch's law: $Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}$; NCERT values (S cm² mol⁻¹): $H^{+}$ 349.6, $N a^{+}$ 50.1, $K^{+}$ 73.5, $C a^{2 +}$ 119.0, $M g^{2 +}$ 106.0, $O H^{-}$ 199.1, $C l^{-}$ 76.3, $B r^{-}$ 78.1, $C H_{3} C O O^{-}$ 40.9, $S O_{4}^{2 -}$ 160.0.
Degree of dissociation and K_a: $α = \frac{Λ _{m}}{Λ _{m}^{\circ}}, K_{a} = \frac{c α ^{2}}{1 - α}$; For weak electrolytes only.
Λ° of a weak acid from strong electrolytes: $Λ_{C H_{3} C O O H}^{\circ} = Λ_{H C l}^{\circ} + Λ_{C H_{3} C O O N a}^{\circ} - Λ_{N a C l}^{\circ}$; Pick three strong electrolytes so the unwanted ions cancel.

How to solve Conductivity, Molar Conductivity and Kohlrausch's Law questions

Conductance numericals follow a chain: resistance → conductivity → molar conductivity → (for weak electrolytes) degree of dissociation → $K_{a}$ . Identify where the question starts and where it stops.

1If a standard KCl solution is given, find the cell constant first: $G^{*} = κ_{K C l} \times R_{K C l}$ .
2Find the conductivity of the unknown solution: $κ = G^{*} / R$ . Keep units S cm⁻¹.
3Convert to molar conductivity: $Λ_{m} = \frac{κ \times 1000}{c}$ with $c$ in mol L⁻¹.
4For $Λ_{m}^{\circ}$ , add the ionic values with the right multipliers (Kohlrausch), or combine strong-electrolyte $Λ_{m}^{\circ}$ values so that spectator ions cancel.
5For a weak electrolyte: $α = Λ_{m} / Λ_{m}^{\circ}$ , then $K_{a} = \frac{c α ^{2}}{1 - α}$ .
6For a strong electrolyte with two data points, solve $Λ_{m} = Λ_{m}^{\circ} - A c$ as two simultaneous equations.
7State units at every step: S cm⁻¹ for $κ$ , S cm² mol⁻¹ for $Λ_{m}$ .

Conductivity, Molar Conductivity and Kohlrausch's Law in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Using molar conductivity to estimate concentration

Suppose a gardener dissolves potassium chloride fertiliser in water and wants to check its concentration with a handheld conductivity meter. The meter reads $1.5 \times 1 0^{- 3}$ S cm⁻¹. The solution is dilute, so its molar conductivity can be taken as roughly equal to the limiting value.

Using $λ^{\circ} (K^{+}) = 73.5$ and $λ^{\circ} (C l^{-}) = 76.3$ S cm² mol⁻¹, estimate the concentration of KCl.

Try it first, then show the solutionSolution

Kohlrausch: $Λ_{m}^{\circ} (K C l) = 73.5 + 76.3 = 149.8$ S cm² mol⁻¹.
Take $Λ_{m} \approx Λ_{m}^{\circ}$ for a dilute strong electrolyte.
$c = \frac{κ \times 1000}{Λ _{m}} = \frac{1.5 \times 1 0 ^{- 3} \times 1000}{149.8} \approx 0.010$ mol L⁻¹.
The real $Λ_{m}$ is slightly below $Λ_{m}^{\circ}$ , so the true concentration is a little higher than this estimate.

Answer: About 0.010 mol L⁻¹ (a slight underestimate).

Memory hook: A conductivity meter counts moving ions; divide by what one mole of ions can carry to get concentration.

Tricks to remember Conductivity, Molar Conductivity and Kohlrausch's Law

Short memory aids. Say each one aloud once, then use it on a question the same day.

κ counts ions per cube, Λ counts ions per mole
Dilute it: fewer ions per cube (κ down) but each mole's ions roam freer (Λ up).
Use it when: Answers every 'how do κ and Λm change on dilution' question.
1000 over c
Λ = κ × 1000 / c: the 1000 converts litres to cm³.
Use it when: Getting the units S cm² mol⁻¹ right.
Add the two you need, subtract the one you don't
For $Λ^{\circ} (H A)$ : (H with anything) + (A with anything) − (the two 'anythings' together).
Use it when: Choosing strong electrolytes in Kohlrausch questions.

Conductivity, Molar Conductivity and Kohlrausch's Law: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Units of molar conductivity · 1 mark · MCQBoard level

The SI unit of molar conductivity is:
(a) S m⁻¹ (b) S m² mol⁻¹ (c) S m mol⁻¹ (d) S⁻¹ m² mol

Try it first, then show the solutionSolution

$Λ_{m} = κ / c$ : (S m⁻¹)/(mol m⁻³) = S m² mol⁻¹.

Answer: (b) S m² mol⁻¹.

Example 2: Molar conductivity from conductivity · 2 marks · Very short answerBoard level

The conductivity of a 0.025 M NaCl solution at 298 K is $2.96 \times 1 0^{- 3}$ S cm⁻¹. Calculate its molar conductivity.

Try it first, then show the solutionSolution

$Λ_{m} = \frac{κ \times 1000}{c}$ (½ mark).
$= \frac{2.96 \times 1 0 ^{- 3} \times 1000}{0.025}$ (½ mark).
$= 118.4$ S cm² mol⁻¹ (1 mark).

Answer: 118.4 S cm² mol⁻¹.

Example 3: Cell constant, then molar conductivity · 3 marks · Short answerHarder board level

A conductivity cell filled with 0.10 M KCl solution (conductivity 0.0129 S cm⁻¹) shows a resistance of 85 Ω at 298 K. The same cell filled with 0.050 M NaCl solution shows 160 Ω. Calculate (a) the cell constant, (b) the conductivity and (c) the molar conductivity of the NaCl solution.

Try it first, then show the solutionSolution

(a) $G^{*} = κ \times R = 0.0129 \times 85 = 1.097$ cm⁻¹ (1 mark).
(b) $κ = \frac{G ^{*}}{R} = \frac{1.097}{160} = 6.85 \times 1 0^{- 3}$ S cm⁻¹ (1 mark).
(c) $Λ_{m} = \frac{6.85 \times 1 0 ^{- 3} \times 1000}{0.050} = 137$ S cm² mol⁻¹ (1 mark).

Answer: (a) 1.10 cm⁻¹. (b) $6.85 \times 1 0^{- 3}$ S cm⁻¹. (c) About 137 S cm² mol⁻¹.

Example 4: Degree of dissociation and K_a of ethanoic acid · 3 marks · Short answerHarder board level

The conductivity of 0.0050 M ethanoic acid at 298 K is $1.15 \times 1 0^{- 4}$ S cm⁻¹. Calculate its degree of dissociation and dissociation constant. ( $λ^{\circ} (H^{+}) = 349.6$ , $λ^{\circ} (C H_{3} C O O^{-}) = 40.9$ S cm² mol⁻¹)

Try it first, then show the solutionSolution

$Λ_{m} = \frac{1.15 \times 1 0 ^{- 4} \times 1000}{0.0050} = 23.0$ S cm² mol⁻¹ (½ mark).
$Λ_{m}^{\circ} = 349.6 + 40.9 = 390.5$ S cm² mol⁻¹ (½ mark).
$α = \frac{23.0}{390.5} = 0.0589$ (1 mark).
$K_{a} = \frac{c α ^{2}}{1 - α} = \frac{0.0050 \times ( 0.0589 ) ^{2}}{1 - 0.0589} = 1.84 \times 1 0^{- 5}$ (1 mark).

Answer: $α \approx 0.059$ (5.9%); $K_{a} \approx 1.8 \times 1 0^{- 5}$ .

Example 5: Extracting Λ° and A for a strong electrolyte · 3 marks · Short answerHOTS / competency

For a strong 1-1 electrolyte at 298 K, $Λ_{m}$ is 124.0 S cm² mol⁻¹ at 0.010 M and 118.0 S cm² mol⁻¹ at 0.040 M. (a) Find $Λ_{m}^{\circ}$ and the constant $A$ . (b) Predict $Λ_{m}$ at 0.0025 M. (c) Why can the same method not be used for ethanoic acid?

Try it first, then show the solutionSolution

$0.010 = 0.10$ , $0.040 = 0.20$ . Equations: $124.0 = Λ_{m}^{\circ} - 0.10 A$ and $118.0 = Λ_{m}^{\circ} - 0.20 A$ (½ mark).
Subtracting: $0.10 A = 6.0$ , $A = 60$ ; $Λ_{m}^{\circ} = 124.0 + 6.0 = 130.0$ S cm² mol⁻¹ (1 mark).
(b) $0.0025 = 0.05$ : $Λ_{m} = 130.0 - 60 \times 0.05 = 127.0$ S cm² mol⁻¹ (½ mark).
(c) A weak electrolyte's $Λ_{m}$ does not vary linearly with $c$ ; it rises steeply at very low concentration as $α$ increases, so extrapolation to $c = 0$ is unreliable. Kohlrausch's law is used instead (1 mark).

Answer: (a) $Λ_{m}^{\circ} = 130.0$ S cm² mol⁻¹, $A = 60$ . (b) 127.0 S cm² mol⁻¹. (c) Its plot is not linear, so $Λ_{m}^{\circ}$ comes from Kohlrausch's law.

Where students lose marks in Conductivity, Molar Conductivity and Kohlrausch's Law

The slips examiners see most often, and what to write instead.

Mistake: Forgetting the factor 1000
Mixing mol L⁻¹ with S cm⁻¹.
Wrong working
$Λ_{m} = \frac{2.96 \times 1 0 ^{- 3}}{0.025} = 0.118$
Right working
$Λ_{m} = \frac{2.96 \times 1 0 ^{- 3} \times 1000}{0.025} = 118.4$ S cm² mol⁻¹
Do this: Check that $Λ_{m}$ for a common salt comes out roughly 50-450 S cm² mol⁻¹.
Usually costs: 1 mark (the final answer).
Mistake: Saying conductivity increases on dilution
Conductivity and molar conductivity are confused.
Do this: Conductivity is per volume (falls); molar conductivity is per mole (rises).
Usually costs: 1 mark in theory questions.
Mistake: Ignoring the number of ions in Kohlrausch's law
$C a C l_{2}$ treated like NaCl.
Wrong working
$Λ_{m}^{\circ} (C a C l_{2}) = 119.0 + 76.3 = 195.3$
Right working
$Λ_{m}^{\circ} (C a C l_{2}) = 119.0 + 2 (76.3) = 271.6$ S cm² mol⁻¹
Do this: Multiply each $λ^{\circ}$ by its subscript in the formula.
Usually costs: 1 mark.
Mistake: Using $α = Λ_{m} / Λ_{m}^{\circ}$ for a strong electrolyte
The formula is applied to KCl.
Do this: This ratio gives the degree of dissociation only for weak electrolytes; strong electrolytes are fully ionised and the drop in $Λ_{m}$ is due to inter-ionic attraction.
Usually costs: 1 mark in reasoning questions.

Exam tips

Draw or describe the two graphs ( $Λ_{m}$ against $c$ ) side by side: a gently falling straight line for KCl and a steep curve for ethanoic acid.
State Kohlrausch's law in words before using it; the statement often carries one mark.
Write units in every line; the board marking scheme deducts for missing units in final answers.
When given resistance of a standard KCl solution, always compute the cell constant first.
For $K_{a}$ , keep $(1 - α)$ in the denominator unless $α$ is very small and the question allows the approximation.

Practice questions on Conductivity, Molar Conductivity and Kohlrausch's Law

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
On diluting an aqueous solution of KCl:
(a) conductivity increases and molar conductivity decreases
(b) both conductivity and molar conductivity decrease
(c) conductivity decreases and molar conductivity increases
(d) both increase
Hint
Ions per cm³ versus ions per mole.
Answer
(c).
Q2 · 3 marks · Short answerBoard level
Define conductivity and molar conductivity of an electrolyte solution. How does each change with dilution? Give reasons.
Hint
Conductivity is per unit volume; molar conductivity is per mole.
Answer
Conductivity is the conductance of a solution of 1 cm length and 1 cm² cross-section (1 cm³); molar conductivity is the conductivity divided by concentration, i.e. the conducting power of all ions from 1 mol of electrolyte. On dilution conductivity falls because there are fewer ions per unit volume; molar conductivity rises because the volume containing one mole increases and ions move more freely (and, for weak electrolytes, ionisation increases).
Q3 · 2 marks · Very short answerHarder board level
Using the values $λ^{\circ} (K^{+}) = 73.5$ , $λ^{\circ} (N a^{+}) = 50.1$ , $λ^{\circ} (B r^{-}) = 78.1$ and $λ^{\circ} (S O_{4}^{2 -}) = 160.0$ S cm² mol⁻¹, calculate $Λ_{m}^{\circ}$ for $K_{2} S O_{4}$ and NaBr.
Hint
Multiply by the number of each ion in the formula.
Answer
$Λ_{m}^{\circ} (K_{2} S O_{4}) = 2 (73.5) + 160.0 = 307.0$ S cm² mol⁻¹; $Λ_{m}^{\circ} (N a B r) = 50.1 + 78.1 = 128.2$ S cm² mol⁻¹.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $Λ_{m}^{\circ}$ of $C a C l_{2}$ equals $λ^{\circ} (C a^{2 +}) + λ^{\circ} (C l^{-})$ .
Reason (R): At infinite dilution each ion migrates independently of the other ions present.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Count the chloride ions in the formula.
Answer
(d). R is Kohlrausch's law and is true, but A is false: $Λ_{m}^{\circ} (C a C l_{2}) = λ^{\circ} (C a^{2 +}) + 2 λ^{\circ} (C l^{-})$ .
Q5 · 2 marks · Very short answerHarder board level
A conductivity cell of cell constant 0.85 cm⁻¹ is filled with a 0.020 M solution of an electrolyte. Its resistance is 520 Ω. Calculate the conductivity and molar conductivity of the solution.
Hint
$κ = G^{*} / R$ .
Answer
$κ = \frac{0.85}{520} = 1.63 \times 1 0^{- 3}$ S cm⁻¹; $Λ_{m} = \frac{1.63 \times 1 0 ^{- 3} \times 1000}{0.020} = 81.7$ S cm² mol⁻¹.
Q6 · 2 marks · Very short answerHarder board level
Given $Λ_{m}^{\circ}$ values (S cm² mol⁻¹): $N H_{4} C l$ = 149.7, NaOH = 249.2 and NaCl = 126.4. Calculate $Λ_{m}^{\circ}$ for $N H_{4} O H$ and state the law used.
Hint
Add the two containing $N H_{4}^{+}$ and $O H^{-}$ , subtract NaCl.
Answer
Kohlrausch's law: $Λ_{m}^{\circ} (N H_{4} O H) = 149.7 + 249.2 - 126.4 = 272.5$ S cm² mol⁻¹.
Q7 · 3 marks · Short answerHOTS / competency
The molar conductivity of 0.010 M methanoic acid (HCOOH) at 298 K is 50.5 S cm² mol⁻¹. Calculate its degree of dissociation and dissociation constant. ( $λ^{\circ} (H^{+}) = 349.6$ , $λ^{\circ} (H C O O^{-}) = 54.6$ S cm² mol⁻¹) Would $α$ be larger or smaller at 0.0010 M? Why?
Hint
$Λ_{m}^{\circ} = 404.2$ .
Answer
$α = \frac{50.5}{404.2} = 0.125$ ; $K_{a} = \frac{0.010 \times ( 0.125 ) ^{2}}{1 - 0.125} = 1.78 \times 1 0^{- 4}$ . At 0.0010 M $α$ is larger: $K_{a}$ is constant, so a lower $c$ needs a larger $α$ (dilution favours ionisation).
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a water-testing lab checks the purity of distilled water by conductivity. A cell of cell constant 1.0 cm⁻¹ gives 2500 Ω for a sample of tap water and a very high resistance for freshly distilled water. The meter uses an alternating current source.
(a) Calculate the conductivity of the tap water. (1 mark)
(b) Why is the conductivity of pure water so low? (1 mark)
(c) Why is alternating current, not direct current, used in the measurement? (1 mark)
(d) Will the reading rise or fall if the tap water sample is diluted with distilled water? Explain. (1 mark)
Hint
$κ = G^{*} / R$ .
Answer
(a) $κ = 1.0/2500 = 4.0 \times 1 0^{- 4}$ S cm⁻¹. (b) Pure water ionises only very slightly, so it has very few ions. (c) Direct current would electrolyse the solution at the electrodes and change its composition and resistance. (d) Conductivity falls (resistance rises), because the number of ions per unit volume decreases.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Formula for cell constant?Show answer
$G^{*} = l / A = κ \times R$ .
2.Unit of conductivity in CGS units used in NCERT problems?Show answer
S cm⁻¹.
3.State Kohlrausch's law.Show answer
Limiting molar conductivity of an electrolyte equals the sum of the limiting molar conductivities of its cations and anions, each multiplied by its number.
4.How is $α$ found from conductance data?Show answer
$α = Λ_{m} / Λ_{m}^{\circ}$ .
5.Which electrolyte's $Λ_{m}^{\circ}$ can be found by extrapolation?Show answer
A strong electrolyte, from the straight line of $Λ_{m}$ against $c$ .
6. $λ^{\circ}$ of $H^{+}$ from NCERT?Show answer
349.6 S cm² mol⁻¹.

Conductivity, Molar Conductivity and Kohlrausch's Law: frequently asked questions

What is molar conductivity in Class 12 Chemistry?

It is the conductivity of a solution divided by its molar concentration, \Lambda_m = \frac{\kappa \times 1000}{c} S cm² mol⁻¹, i.e. the conducting power of all the ions produced by one mole of electrolyte.

Why does molar conductivity increase with dilution?

For strong electrolytes, ions are farther apart and inter-ionic attraction falls, so they move faster. For weak electrolytes, the degree of ionisation also increases. In both, the volume holding one mole grows.

What is Kohlrausch's law used for?

To find the limiting molar conductivity of weak electrolytes, and from it their degree of dissociation and dissociation constant.

Why does conductivity decrease on dilution?

Conductivity measures ions in a fixed volume (1 cm³). Diluting reduces the number of ions in that volume.

Why is AC used in conductivity measurement?

Direct current would cause electrolysis at the electrodes, changing the solution and its resistance during the measurement.