Molar Conductivity and Kohlrausch's Law — Calculations

By Aniket Bhardwaj · 14 September 2026 · Calculator/Formula Guide

Conductivity questions lose marks for one reason above all others: units. The physics is straightforward — ions carry current, more ions carry more current — but the numbers travel between S cm⁻¹, S m⁻¹, S cm² mol⁻¹ and S m² mol⁻¹, and a factor of 1000 sits in the middle of the standard formula. This page fixes the units first, then works through molar conductivity and both statements of Kohlrausch's law.

Three quantities, in order

Conductance G = 1/R (unit: siemens, S = Ω⁻¹)

Conductivity κ = G × (cell constant) (unit: S cm⁻¹ or S m⁻¹)

Molar conductivity Λ_m = κ ÷ c

Conductance belongs to the particular cell you used — change the electrodes and it changes. Conductivity removes the geometry and describes the solution: it is the conductance of a 1 cm cube (or 1 m cube in SI) of that solution. Molar conductivity goes one step further and removes the concentration, so it describes how well one mole of the electrolyte conducts. That is the quantity you can compare fairly between solutions.

Where the 1000 comes from

In practical work κ is measured in S cm⁻¹ but concentration is quoted in mol L⁻¹, and a litre is 1000 cm³. Converting c into mol cm⁻³ means dividing it by 1000, so:

Λ_m (S cm² mol⁻¹) = κ (S cm⁻¹) × 1000 ÷ c (mol L⁻¹)

In strict SI there is no 1000 at all: with κ in S m⁻¹ and c in mol m⁻³, Λ_m comes out in S m² mol⁻¹ straight away. The 1000 is not a piece of chemistry — it is the price of using centimetres and litres in the same equation.

Quantity	Practical unit	SI unit	Conversion
Conductivity κ	S cm⁻¹	S m⁻¹	1 S cm⁻¹ = 100 S m⁻¹
Molar conductivity Λ_m	S cm² mol⁻¹	S m² mol⁻¹	1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹
Concentration c	mol L⁻¹	mol m⁻³	1 mol L⁻¹ = 1000 mol m⁻³

Worked example 1 — molar conductivity from conductivity

A 0.20 mol L⁻¹ KCl solution has κ = 0.0248 S cm⁻¹. Find Λ_m.

Λ_m = 0.0248 × 1000 ÷ 0.20 = 24.8 ÷ 0.20 = 124 S cm² mol⁻¹

In SI: 124 S cm² mol⁻¹ = 124 × 10⁻⁴ = 1.24 × 10⁻² S m² mol⁻¹. Both are the same physical quantity; write whichever unit your question uses.

Worked example 2 — starting from a resistance reading

A conductivity cell of cell constant 1.29 cm⁻¹ contains a 0.010 mol L⁻¹ solution and reads 1000 Ω. Find κ and Λ_m.

Step 1 — conductance. G = 1 ÷ R = 1 ÷ 1000 = 1.00 × 10⁻³ S

Step 2 — conductivity.
κ = G × cell constant = 1.00 × 10⁻³ × 1.29 = 1.29 × 10⁻³ S cm⁻¹

Step 3 — molar conductivity.
Λ_m = 1.29 × 10⁻³ × 1000 ÷ 0.010 = 1.29 ÷ 0.010 = 129 S cm² mol⁻¹

What happens on dilution — and the trap inside it

Dilute a solution and two things move in opposite directions:

κ falls. There are fewer ions in each cubic centimetre, so a fixed volume of solution conducts less.
Λ_m rises. Λ_m already divides by concentration. What is left is how freely each mole of ions moves, and ions interfere with one another less when they are further apart.

Λ_m approaches a limit at infinite dilution, written Λ°_m and called the limiting molar conductivity. For a strong electrolyte the approach is described by

Λ_m = Λ°_m − A√c

so a plot of Λ_m against √c is a straight line whose intercept is Λ°_m. This relation is itself often called "Kohlrausch's law", which is one reason the topic confuses students — the name is used for two different statements. The square-root relation above is one; the law of independent migration below is the other. When a question says "Kohlrausch's law", read the rest of the sentence to see which is meant.

For a weak electrolyte the extrapolation fails completely. Λ_m climbs very steeply near zero concentration, because dilution is shifting a dissociation equilibrium rather than merely spacing out ions that were already free. No straight line exists to extrapolate — which is exactly the problem the next section solves.

Kohlrausch's law of independent migration

Λ°_m = ν₊ λ°₊ + ν₋ λ°₋

At infinite dilution each ion contributes its own fixed amount λ°, independent of the other ion it arrived with, and ν₊ and ν₋ are the numbers of each ion in one formula unit. Cl⁻ contributes the same whether it came from HCl, NaCl or CaCl₂.

Example 3 — a limiting molar conductivity from ionic values. Using λ°(Na⁺) = 50.1 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹ from a standard 298 K data table:

Λ°(NaCl) = 50.1 + 76.3 = 126.4 S cm² mol⁻¹

For CaCl₂ there are two chloride ions per formula unit, and the stoichiometric coefficient must not be forgotten. With λ°(Ca²⁺) = 119.0 S cm² mol⁻¹:

Λ°(CaCl₂) = 119.0 + (2 × 76.3) = 119.0 + 152.6 = 271.6 S cm² mol⁻¹

A convention warning. Some data tables list ionic conductivities per unit of charge, writing λ°(½Ca²⁺) = 59.5 instead of λ°(Ca²⁺) = 119.0. Both tables are correct; they define the "mole" of ion differently. Check the heading of the table before you add, or your answer for any multivalent ion will be out by a factor equal to its charge.

Worked example 4 — the classic weak-electrolyte problem

Λ° for acetic acid cannot be measured by extrapolation, but it can be assembled from three strong electrolytes that between them contain the right ions:

Λ°(CH₃COOH) = Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl)

Why this works — write out the ions:
(H⁺ + Cl⁻) + (CH₃COO⁻ + Na⁺) − (Na⁺ + Cl⁻) = H⁺ + CH₃COO⁻ ✔
The sodium and chloride contributions cancel exactly, which is only allowed because each ion contributes independently.

Using standard 298 K values Λ°(HCl) = 426.16, Λ°(CH₃COONa) = 91.0 and Λ°(NaCl) = 126.4 S cm² mol⁻¹:

Λ°(CH₃COOH) = 426.16 + 91.0 − 126.4 = 390.8 S cm² mol⁻¹

That number is the gateway to the rest of the chapter. Once Λ° is known, the degree of dissociation of the weak acid at any concentration follows from α = Λ_m ÷ Λ°_m, and from α you can reach the dissociation constant.

Common mistakes that cost marks

Dropping the 1000. With κ in S cm⁻¹ and c in mol L⁻¹ the factor is compulsory. Omit it and every answer is 1000 times too small.
Mixing SI and practical units in one line. Either work entirely in S cm⁻¹ and mol L⁻¹, or entirely in S m⁻¹ and mol m⁻³. Half-converted working is the single biggest source of wrong answers here.
Saying conductivity increases on dilution. κ decreases; Λ_m increases. Name the quantity before you state the trend.
Extrapolating a weak electrolyte to zero concentration. The Λ_m against √c line is for strong electrolytes only.
Forgetting ν₊ and ν₋. Λ°(CaCl₂) needs two chlorides, and Λ°(Al₂(SO₄)₃) needs two aluminium and three sulphate contributions.
Confusing cell constant with conductance. The cell constant has units of cm⁻¹ and comes from the cell's geometry; it multiplies G to give κ.

Where this appears in exams

Level	Typical question
CBSE/ICSE Class 12	Λ_m from κ and c; Λ° of a weak acid by the three-electrolyte method
School practical work	Measure resistance of KCl and a test solution, calculate κ and Λ_m
IIT-JAM / CUET-PG	Λ_m against √c plots, degree of dissociation, Ostwald's dilution law
GATE / CSIR-NET	Ionic mobilities, transport numbers and where the Debye–Hückel–Onsager treatment applies

There is no dedicated conductivity tool in the suite yet, so this is one of the calculations to do on the Scientific Calculator — and it is honest to say so rather than send you to the wrong screen. Open the suite for the scientific keypad, and use the Concentration tool alongside it when a question makes you find c first.

Open the ABC Chemistry Calculator Suite →

If a question gives you grams per litre instead of molarity, the Molar Mass calculator gets you to mol L⁻¹ in one step.

Electrochemistry rewards students who get the units right the first time. ABC Chemistry teaches Class 11–12 chemistry at its Gurugram centre and in online batches across India, with home tuition available in Delhi-NCR — abcchemistry.in.