ChemistryISC Class 12 · Electrochemistry

Conductivity, Molar Conductivity and Kohlrausch's Law: ISC Class 12

Updated 15 October 2026 · Fact-checked

Specific conductivity κ is the conductance of a 1 cm cube of solution; molar conductivity Λm = κ × 1000 / M. On dilution κ falls but Λm rises. Kohlrausch's law says that at infinite dilution each ion contributes a fixed amount, so Λ°m = ν₊λ°₊ + ν₋λ°₋, which gives Λ° and α for weak electrolytes.

Understand Conductivity, Molar Conductivity and Kohlrausch's Law

Electrolytic conduction. In metals, electrons carry the current and no chemical change occurs; resistance rises with temperature. In electrolyte solutions, ions carry the current, matter moves, chemical changes occur at the electrodes, and conductance usually rises with temperature because ions move faster.

Resistance, conductance and the cell constant. A solution between two parallel electrodes of area $A$ separated by $l$ has resistance $R = ρ \frac{l}{A}$ , where $ρ$ is the resistivity. Conductance $G = \frac{1}{R}$ (unit siemens, S). Specific conductivity (conductivity) $κ = \frac{1}{ρ}$ is the conductance of a solution placed between electrodes 1 cm² in area and 1 cm apart, i.e. of a 1 cm cube; unit S cm⁻¹. So $κ = \frac{1}{R} \cdot \frac{l}{A} = \frac{G ^{*}}{R}$ , where $G^{*} = \frac{l}{A}$ is the cell constant (cm⁻¹). It is found by measuring $R$ for a KCl solution of known κ: $G^{*} = κ R$ .

Molar and equivalent conductivity. κ depends on how many ions are in each cubic centimetre, so it is not a fair way to compare electrolytes. Molar conductivity $Λ_{m}$ is the conductance of all the ions from one mole of electrolyte held between electrodes 1 cm apart: $Λ_{m} = \frac{κ \times 1000}{M}$ with $M$ in mol L⁻¹, unit S cm² mol⁻¹. ISC also uses equivalent conductivity $Λ_{e q} = \frac{κ \times 1000}{N}$ (normality $N$ ), unit S cm² eq⁻¹. For CuSO₄, $N = 2 M$ , so $Λ_{e q} = \frac{1}{2} Λ_{m}$ .

Effect of dilution. On dilution, κ decreases because the number of ions per unit volume falls. Λm increases, because $Λ_{m} = κV$ where $V$ is the volume holding one mole: $V$ grows faster than κ falls. For strong electrolytes (KCl, NaCl, HCl) Λm rises only slightly, because they are already fully ionised and dilution only reduces interionic attraction; a plot of Λm against $c$ is nearly a straight line, $Λ_{m} = Λ_{m}^{\circ} - A c$ (Debye-Hückel-Onsager), and extrapolating to $c = 0$ gives $Λ_{m}^{\circ}$ , the limiting molar conductivity. For weak electrolytes (ethanoic acid, NH₄OH) Λm is small at moderate concentrations and rises steeply on dilution because the degree of ionisation increases. The curve shoots up near zero concentration, so Λ°m cannot be found by extrapolation.

Kohlrausch's law of independent migration of ions. At infinite dilution the ions are so far apart that each moves independently, so the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its ions: $Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}$ , where $ν$ is the number of each ion in the formula. Kohlrausch noticed, for example, that the difference between the Λ° values of a sodium and a potassium salt is the same whatever the common anion.

Applications. (i) Λ° of a weak electrolyte by combining strong electrolytes: $Λ^{\circ} (C H_{3} C O O H) = Λ^{\circ} (C H_{3} C O O N a) + Λ^{\circ} (H C l) - Λ^{\circ} (N a C l)$ . (ii) Degree of dissociation of a weak electrolyte: $α = \frac{Λ _{m}}{Λ _{m}^{\circ}}$ . (iii) Its dissociation constant: $K_{a} = \frac{c α ^{2}}{1 - α}$ . Typical λ° values at 298 K (S cm² mol⁻¹): H⁺ 349.6, OH⁻ 199.1, K⁺ 73.5, Na⁺ 50.1, Ca²⁺ 119.0, Cl⁻ 76.3, Br⁻ 78.1, CH₃COO⁻ 40.9, SO₄²⁻ 160.0. H⁺ and OH⁻ are exceptionally high.

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

Cell constant and conductivity: $G^{*} = \frac{l}{A}, κ = \frac{G ^{*}}{R} = G G^{*}$; $G^{*}$ in cm⁻¹, κ in S cm⁻¹. Calibrate with KCl: $G^{*} = κ_{K C l} R_{K C l}$ .
Molar conductivity: $Λ_{m} = \frac{κ \times 1000}{M}$; $M$ in mol L⁻¹; unit S cm² mol⁻¹ (1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹).
Equivalent conductivity: $Λ_{e q} = \frac{κ \times 1000}{N}$; $N$ = normality; unit S cm² eq⁻¹.
Kohlrausch's law: $Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}$; Example: $Λ^{\circ} (C a C l_{2}) = λ_{C a^{2 +}}^{\circ} + 2 λ_{C l^{-}}^{\circ}$ .
Degree and constant of dissociation: $α = \frac{Λ _{m}}{Λ _{m}^{\circ}}, K_{a} = \frac{c α ^{2}}{1 - α}$; For weak electrolytes only.
Strong electrolytes on dilution: $Λ_{m} = Λ_{m}^{\circ} - A c$; Straight line on a Λm against √c plot; intercept gives Λ°m.

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

ISC conductivity numericals are a chain: resistance to κ, κ to Λm, Kohlrausch to Λ°m, then α. Questions stop at any link.

1If a standard KCl solution is given, find the cell constant first: $G^{*} = κ_{K C l} \times R_{K C l}$ .
2For the test solution, $κ = \frac{G ^{*}}{R}$ ; keep at least four significant figures.
3Convert to molar conductivity with $Λ_{m} = \frac{κ \times 1000}{M}$ (or equivalent conductivity with $N$ ). Check the unit: S cm² mol⁻¹.
4For Λ°m of a salt, write the formula and multiply each λ° by the number of that ion; for a weak electrolyte, add and subtract strong-electrolyte Λ° values so that only the wanted ions remain.
5For a weak electrolyte, $α = \frac{Λ _{m}}{Λ _{m}^{\circ}}$ ; then $K_{a} = \frac{c α ^{2}}{1 - α}$ if asked.
6In reasoning questions about dilution, say 'number of ions per unit volume' for κ and 'volume containing one mole' for Λm.

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.

Conductivity measures dissolved ions

Suppose an aquarium keeper fills a tank with purified water whose conductivity is 2 × 10⁻⁶ S cm⁻¹, then adds a little table salt for a freshwater fish. A conductivity meter now reads 1.20 × 10⁻⁴ S cm⁻¹ at 298 K. Assume the salt is NaCl, the solution is dilute enough that $Λ_{m} \approx Λ_{m}^{\circ}$ , and use λ° Na⁺ = 50.1 and Cl⁻ = 76.3 S cm² mol⁻¹.

Estimate the concentration of NaCl in mol L⁻¹ and in g L⁻¹.

Try it first, then show the solutionSolution

Conductivity due to the salt $= 1.20 \times 1 0^{- 4} - 0.02 \times 1 0^{- 4} = 1.18 \times 1 0^{- 4}$ S cm⁻¹.
$Λ_{m}^{\circ} (N a C l) = 50.1 + 76.3 = 126.4$ S cm² mol⁻¹.
$M = \frac{κ \times 1000}{Λ _{m}} = \frac{1.18 \times 1 0 ^{- 4} \times 1000}{126.4} = 9.34 \times 1 0^{- 4}$ mol L⁻¹.
Mass per litre $= 9.34 \times 1 0^{- 4} \times 58.5 = 0.0546$ g L⁻¹.

Answer: About $9.3 \times 1 0^{- 4}$ mol L⁻¹, i.e. roughly 0.055 g of NaCl per litre.

Memory hook: A conductivity meter is an ion counter: κ × 1000 / Λ gives the 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 crowd, Λ counts a mole
κ is about how crowded a 1 cm cube is with ions, so it drops on dilution. Λm follows one whole mole, which spreads out and moves more freely, so it rises.
Use it when: Any 'why does conductivity decrease but molar conductivity increase' question.
Kappa times a thousand over M
$Λ_{m} = \frac{1000 κ}{M}$ : 'a thousand kappas per mole'.
Use it when: The core conversion in every numerical.
Add the wanted, cancel the unwanted
For Λ°(weak acid HA): HCl + NaA − NaCl. Na⁺ and Cl⁻ appear once with + and once with −, leaving H⁺ + A⁻.
Use it when: Kohlrausch combination problems.

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: Dilution and a weak electrolyte · 1 mark · MCQBoard level

When an aqueous solution of ethanoic acid is diluted, its molar conductivity:
(a) decreases slowly
(b) stays constant
(c) increases sharply, especially at low concentration
(d) increases slowly in a straight line with √c

Try it first, then show the solutionSolution

Ethanoic acid is a weak electrolyte; dilution increases its degree of ionisation.
So Λm rises steeply at low concentration; the straight-line rise is typical of strong electrolytes.

Answer: (c).

Example 2: Molar conductivity from conductivity · 2 marks · Short answerBoard level

The conductivity of a 0.010 M solution of KCl at 298 K is $1.41 \times 1 0^{- 3}$ S cm⁻¹. Calculate its molar conductivity.

Try it first, then show the solutionSolution

$Λ_{m} = \frac{κ \times 1000}{M}$ (1 mark).
$Λ_{m} = \frac{1.41 \times 1 0 ^{- 3} \times 1000}{0.010} = 141$ S cm² mol⁻¹ (1 mark).

Answer: 141 S cm² mol⁻¹.

Example 3: Cell constant, then a test solution · 3 marks · Short answerHarder board level

A conductivity cell filled with a standard KCl solution of conductivity 0.0140 S cm⁻¹ has a resistance of 90.0 Ω at 298 K. The same cell filled with 0.050 M NaCl solution has a resistance of 242 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) $G^{*} = κ R = 0.0140 \times 90.0 = 1.26$ cm⁻¹ (1 mark).
(b) $κ = \frac{G ^{*}}{R} = \frac{1.26}{242} = 5.21 \times 1 0^{- 3}$ S cm⁻¹ (1 mark).
(c) $Λ_{m} = \frac{5.21 \times 1 0 ^{- 3} \times 1000}{0.050} = 104$ S cm² mol⁻¹ (1 mark).

Answer: (a) 1.26 cm⁻¹. (b) $5.21 \times 1 0^{- 3}$ S cm⁻¹. (c) About 104 S cm² mol⁻¹.

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

The molar conductivity of 0.0010 M ethanoic acid at 298 K is 48.5 S cm² mol⁻¹. Using λ°(H⁺) = 349.6 and λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹, calculate the degree of dissociation and the dissociation constant of the acid.

Try it first, then show the solutionSolution

Kohlrausch: $Λ_{m}^{\circ} = 349.6 + 40.9 = 390.5$ S cm² mol⁻¹ (1 mark).
$α = \frac{48.5}{390.5} = 0.124$ (1 mark).
$K_{a} = \frac{c α ^{2}}{1 - α} = \frac{0.0010 \times ( 0.124 ) ^{2}}{0.876} = 1.76 \times 1 0^{- 5}$ (1 mark).

Answer: $α \approx 0.124$ (12.4%); $K_{a} \approx 1.76 \times 1 0^{- 5}$ .

Example 5: Kohlrausch and a weak base · 5 marks · Long answerHOTS / competency

(a) Calculate $Λ_{m}^{\circ}$ of Ca(OH)₂ from λ°(Ca²⁺) = 119.0 and λ°(OH⁻) = 199.1 S cm² mol⁻¹. [1]
(b) Given $Λ_{m}^{\circ}$ (NH₄Cl) = 149.7, and using λ°(Na⁺) = 50.1, λ°(Cl⁻) = 76.3 and λ°(OH⁻) = 199.1 S cm² mol⁻¹, calculate $Λ_{m}^{\circ}$ of NH₄OH by combining strong electrolytes. [2]
(c) A 0.020 M NH₄OH solution has a resistance of 1250 Ω in a cell of constant 0.80 cm⁻¹. Find its molar conductivity and degree of dissociation. [2]

Try it first, then show the solutionSolution

(a) $Λ^{\circ} (C a (O H)_{2}) = 119.0 + 2 \times 199.1 = 517.2$ S cm² mol⁻¹ (1 mark).
(b) $Λ^{\circ} (N H_{4} O H) = Λ^{\circ} (N H_{4} C l) + Λ^{\circ} (N a O H) - Λ^{\circ} (N a C l)$ (1 mark).
$Λ^{\circ} (N a O H) = 249.2$ , $Λ^{\circ} (N a C l) = 126.4$ , so $Λ^{\circ} (N H_{4} O H) = 149.7 + 249.2 - 126.4 = 272.5$ S cm² mol⁻¹ (1 mark).
(c) $κ = \frac{0.80}{1250} = 6.40 \times 1 0^{- 4}$ S cm⁻¹; $Λ_{m} = \frac{6.40 \times 1 0 ^{- 4} \times 1000}{0.020} = 32.0$ S cm² mol⁻¹ (1 mark).
$α = \frac{32.0}{272.5} = 0.117$ , about 11.7% dissociated (1 mark).

Answer: (a) 517.2 S cm² mol⁻¹. (b) 272.5 S cm² mol⁻¹. (c) $Λ_{m} = 32.0$ S cm² mol⁻¹; $α \approx 0.117$ .

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

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

Mistake: Leaving out 'per unit volume'
ISC examiners noted answers saying only that the number of ions decreases on dilution (2025).
Wrong working
Conductivity decreases on dilution because the number of ions decreases.
Right working
Conductivity decreases on dilution because the number of ions per unit volume (per cm³) decreases.
Do this: The total number of ions does not fall on dilution; only their concentration does. Say 'per unit volume'.
Usually costs: 1 mark.
Mistake: Ignoring the number of ions in Kohlrausch's law
Examiners reported Λ° of BaCl₂ and Al₂(SO₄)₃ calculated without multiplying by the number of ions (2024).
Wrong working
$Λ^{\circ} (A l_{2} (S O_{4})_{3}) = λ_{A l^{3 +}}^{\circ} + λ_{S O_{4}^{2 -}}^{\circ}$
Right working
$Λ^{\circ} (A l_{2} (S O_{4})_{3}) = 2 λ_{A l^{3 +}}^{\circ} + 3 λ_{S O_{4}^{2 -}}^{\circ}$
Do this: Write the formula, then the coefficient of each ion in front of its λ°.
Usually costs: 1 mark per salt.
Mistake: Rounding too early
Examiners noted that approximate values gave incorrect molar conductance in a 2024 NH₄OH question, and advised four decimal places.
Do this: Carry κ to four significant figures, then round Λm and α at the end.
Usually costs: 1 mark.
Mistake: Wrong formula for degree of dissociation
Examiners reported incorrect α formulas in conductance questions.
Wrong working
$α = \frac{Λ _{m}^{\circ}}{Λ _{m}}$
Right working
$α = \frac{Λ _{m}}{Λ _{m}^{\circ}}$ (always less than 1)
Do this: α is a fraction: the smaller value goes on top.
Usually costs: 1 to 2 marks.

Exam tips

State units at every stage: Ω, cm⁻¹, S cm⁻¹, S cm² mol⁻¹. A wrong unit costs the answer mark.
When asked to 'state Kohlrausch's law', include 'at infinite dilution' and 'independent of the other ion'.
In assertion-reason items on dilution, both the conductivity statement and the 'ions per unit volume' reason are usually true; read carefully before choosing (c).
For Λ° of a weak electrolyte, write the strong electrolytes you combine and show which ions cancel.
Do not use α = Λm/Λ°m for strong electrolytes in ISC answers; say they are fully ionised.

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
The unit of cell constant commonly used in ISC numericals is:
(a) cm
(b) cm⁻¹
(c) S cm⁻¹
(d) S cm² mol⁻¹
Hint
$G^{*} = l / A$ .
Answer
(b) cm⁻¹.
Q2 · 2 marks · Short answerBoard level
Define molar conductivity and equivalent conductivity, and write each in terms of the specific conductivity κ.
Hint
One uses molarity, the other normality.
Answer
Molar conductivity is the conductance of all the ions from one mole of electrolyte placed between electrodes 1 cm apart: $Λ_{m} = \frac{κ \times 1000}{M}$ . Equivalent conductivity is the conductance of all the ions from one gram-equivalent: $Λ_{e q} = \frac{κ \times 1000}{N}$ .
Q3 · 2 marks · Short answerHarder board level
The conductivity of 0.050 M CuSO₄ solution is 0.00590 S cm⁻¹ at 298 K. Calculate its molar conductivity and equivalent conductivity.
Hint
For CuSO₄, normality = 2 × molarity.
Answer
$Λ_{m} = \frac{0.00590 \times 1000}{0.050} = 118$ S cm² mol⁻¹. $N = 0.10$ , so $Λ_{e q} = \frac{0.00590 \times 1000}{0.10} = 59.0$ S cm² eq⁻¹.
Q4 · 3 marks · Short answerHarder board level
Given $Λ_{m}^{\circ}$ (S cm² mol⁻¹): HCl 425.9, KBr 151.6, KCl 149.8. (a) Calculate $Λ_{m}^{\circ}$ of HBr. (b) A 0.010 M HBr solution has $Λ_{m}$ = 412 S cm² mol⁻¹. Find the ratio $Λ_{m} / Λ_{m}^{\circ}$ and explain why it is less than 1 although HBr is a strong acid.
Hint
HBr = HCl + KBr − KCl.
Answer
(a) $425.9 + 151.6 - 149.8 = 427.7$ S cm² mol⁻¹. (b) $\frac{412}{427.7} = 0.963$ . HBr is fully ionised; the ratio is below 1 because interionic attractions slow the ions at this concentration, not because of incomplete dissociation.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): On dilution, the conductivity of an electrolyte solution decreases but its molar conductivity increases.
Reason (R): On dilution, the number of ions per unit volume decreases, while the volume of solution containing one mole of electrolyte increases.
(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
$Λ_{m} = κ \times V$ .
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Sketch, in words, the plots of $Λ_{m}$ against $c$ for KCl and for ethanoic acid. Explain the difference in shape, and why $Λ_{m}^{\circ}$ of ethanoic acid cannot be found from its plot.
Hint
Strong versus weak electrolyte.
Answer
KCl: nearly a straight line with a small negative slope; Λm rises slightly on dilution as interionic attraction falls ( $Λ_{m} = Λ_{m}^{\circ} - A c$ ), and extrapolation to $c = 0$ gives Λ°m. Ethanoic acid: Λm is low and almost flat at higher c, then rises very steeply near zero concentration because the degree of ionisation increases sharply. The curve does not approach the axis linearly, so it cannot be extrapolated; Kohlrausch's law is used instead.
Q7 · 3 marks · Short answerHOTS / competency
Very pure water at 298 K has a conductivity of $5.5 \times 1 0^{- 8}$ S cm⁻¹. Using λ°(H⁺) = 349.6 and λ°(OH⁻) = 199.1 S cm² mol⁻¹, calculate the concentration of H⁺ ions in pure water and the ionic product of water.
Hint
Treat water as a very weak electrolyte at infinite dilution: $[H^{+}] = \frac{1000 κ}{λ _{H^{+}}^{\circ} + λ _{O H^{-}}^{\circ}}$ .
Answer
$Λ^{\circ} = 548.7$ S cm² mol⁻¹. $[H^{+}] = [O H^{-}] = \frac{5.5 \times 1 0 ^{- 8} \times 1000}{548.7} = 1.0 \times 1 0^{- 7}$ M. $K_{w} = (1.0 \times 1 0^{- 7})^{2} = 1.0 \times 1 0^{- 14}$ .
Q8 · 3 marks · Short answerHOTS / competency
0.10 M HCl and 0.10 M ethanoic acid have molar conductivities of 391 and 5.2 S cm² mol⁻¹ at 298 K. Using λ° values H⁺ 349.6, Cl⁻ 76.3, CH₃COO⁻ 40.9 S cm² mol⁻¹: (a) calculate $Λ_{m} / Λ_{m}^{\circ}$ for each; (b) explain why the meaning of this ratio is different for the two acids; (c) which solution would light a bulb in a conductivity tester more brightly, and why?
Hint
The ratio is α only for a weak electrolyte.
Answer
(a) HCl: $\frac{391}{425.9} = 0.918$ ; ethanoic acid: $\frac{5.2}{390.5} = 0.0133$ . (b) For ethanoic acid the ratio is the degree of dissociation (1.3%); HCl is fully ionised, and its ratio below 1 only reflects interionic attraction. (c) HCl: its κ ( $3.91 \times 1 0^{- 2}$ S cm⁻¹) is about 75 times that of ethanoic acid ( $5.2 \times 1 0^{- 4}$ S cm⁻¹), because it supplies far more ions.

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.Unit of specific conductivity?Show answer
S cm⁻¹ (S m⁻¹ in SI).
2.How is the cell constant found in practice?Show answer
By measuring the resistance of a KCl solution of known conductivity: $G^{*} = κ R$ .
3.State Kohlrausch's law.Show answer
At infinite dilution, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each contributing independently.
4.Which two ions have exceptionally high λ°?Show answer
H⁺ (349.6) and OH⁻ (199.1 S cm² mol⁻¹).
5.Relation between Λeq and Λm for CuSO₄?Show answer
$Λ_{e q} = \frac{1}{2} Λ_{m}$ , since normality is twice molarity.
6.Why can Λ°m of ethanoic acid not be found by extrapolation?Show answer
Λm rises steeply near zero concentration, so the curve cannot be extrapolated reliably.

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

Why does conductivity decrease but molar conductivity increase on dilution?

Conductivity depends on ions per cm³, which fall on dilution. Molar conductivity is κ times the volume holding one mole; the volume grows faster than κ falls, and the ions also move more freely.

What is the difference between molar and equivalent conductivity?

Molar conductivity refers to one mole of electrolyte (\kappa \times 1000/M); equivalent conductivity refers to one gram-equivalent (\kappa \times 1000/N).

How is Kohlrausch's law used to find the degree of dissociation?

Λ°m of the weak electrolyte is built from ionic λ° values (or strong-electrolyte Λ° values), and then \alpha = \Lambda_m/\Lambda^\circ_m.

Does the conductivity of an electrolyte change with temperature?

Yes. It usually increases with temperature because ions move faster and, for weak electrolytes, ionisation increases.

What is the cell constant?

The ratio l/A of the distance between the electrodes to their area, in cm⁻¹. Multiplying conductance by it gives conductivity.