State Kohlrausch's law of independent migration of ions. With the help of a curve, explain why it is not easy to determine $\Lambda_m^\circ$ for weak electrolytes by extrapolating the concentration $-$ molar conductivity curve, as it is for strong electrolytes.

State Kohlrausch's law of independent migration of ions. With the help of a curve, explain why it is not easy

Question

\Lambda_m^\circ

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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Concept understanding — Molar Conductivity

From Resistance to Conductance: Flipping the Idea

You already know resistance ( $R$ ) — it tells you how much a material opposes the flow of current. A high resistance means the wire fights the current; a low resistance means it lets current through easily.

Now flip that thought. Instead of asking "how much does it resist?", ask "how easily does it let current flow?" That's exactly what conductance measures.

Note

Conductance ( $G$ ) is the reciprocal of resistance:

$G = \frac{1}{R}$

Unit: siemens (S) — named after Werner von Siemens. 1 S = 1 A/V (ampere per volt).

If a wire has $R = 10\ \Omega$ , its conductance is $G = 0.1\ \text{S}$ . If $R = 0.5\ \Omega$ , $G = 2\ \text{S}$ — it conducts twice as well.

Ohm's Law in Conductance Form

You know $V = IR$ . Rearranging:

I = \frac{V}{R} = G V

So current = conductance × voltage. A high-conductance material draws a large current for the same voltage — it's a "good conductor."

Now, Conductivity: The Material's Intrinsic Property

Resistance depends on two things: the material itself (its "resistivity" $\rho$ ) and the geometry (length $L$ , cross-sectional area $A$ ):

R = \rho \frac{L}{A}

Conductance also depends on geometry. A thicker wire (larger $A$ ) or a shorter wire (smaller $L$ ) has higher conductance. To isolate the material's inherent ability to conduct, we define conductivity ( $\sigma$ ):

\sigma = \frac{1}{\rho}

Conductivity is the reciprocal of resistivity. It tells you how well the material itself conducts, independent of shape and size.

Unit: siemens per metre (S/m).
High $\sigma$ → good conductor (copper: $\approx 5.8 \times 10^7\ \text{S/m}$ ).
Low $\sigma$ → poor conductor / insulator (glass: $\approx 10^{-12}\ \text{S/m}$ ).

Watch out

Don't confuse conductance (property of a specific object, depends on geometry) with conductivity (property of the material, independent of geometry). A short thick copper wire has high conductance; a long thin copper wire has lower conductance — but both have the same conductivity.

The Big Picture in One Table

Quantity	Symbol	Definition	Depends on	Unit
Resistance	$R$	$V/I$	Material + geometry	$\Omega$
Resistivity	$\rho$	$R A / L$	Material only	$\Omega \cdot \text{m}$
Conductance	$G$	$1/R$	Material + geometry	S
Conductivity	$\sigma$	$1/\rho$	Material only	S/m

Intuitive Analogy

Think of a water pipe:

Resistance = how hard it is to push water through (narrow, long pipe).
Conductance = how easily water flows (wide, short pipe). …

Why this formula?

Conductance and Conductivity: Why the Formulas Hold

Let's build this from first principles — understanding the why before the what.

1. The Core Idea: How Easily Does Current Flow?

Think of a conductor (like a copper wire). When you apply a voltage across it, electrons drift through the material. Two questions arise:

How much current flows for a given voltage? → This is conductance ( $G$ ).
How well does the material itself allow current? → This is conductivity ( $\sigma$ ).

The key distinction: Conductance depends on the size and shape of the object. Conductivity is an intrinsic property of the material.

2. Ohm's Law in Terms of Conductance

You know Ohm's law:

V = IR

But we can rewrite it as:

I = \frac{V}{R}

Define conductance $G$ as the reciprocal of resistance:

G = \frac{1}{R}

So:

I = G V

Why this makes sense:

A larger $G$ means more current for the same voltage — the conductor "conducts" better.
$G$ has units of siemens (S) = $\text{A/V}$ .

3. From Resistance to Conductivity: The Geometry Factor

Resistance of a uniform conductor depends on:

Length $L$ (longer → more resistance)
Cross-sectional area $A$ (thicker → less resistance)
Material property $\rho$ (resistivity)

The formula:

R = \rho \frac{L}{A}

Now, conductivity $\sigma$ is the reciprocal of resistivity:

\sigma = \frac{1}{\rho}

So:

R = \frac{1}{\sigma} \cdot \frac{L}{A}

Why this form?

If you double the length, electrons have to travel twice as far, colliding more → resistance doubles.
If you double the area, there's twice as many "lanes" for electrons → resistance halves.

4. The Key Formula: Conductance in Terms of Conductivity

Since $G = 1/R$ , we get:

G = \sigma \frac{A}{L}

This is the central relationship. Let's see why it holds:

$\sigma$ tells you how well the material conducts (intrinsic).
$A/L$ tells you how the geometry amplifies or reduces that.

Intuition:

A fat, short wire ( $A$ large, $L$ small) has high conductance.
A thin, long wire ( $A$ small, $L$ large) has low conductance.
A material with high $\sigma$ (like copper) gives higher $G$ than one with low $\sigma$ (like iron), for the same shape.

5. Microscopic Derivation (Why $\sigma$ Exists)

At the microscopic level, conductivity arises from electron motion:

\sigma = n e \mu

Where:

$n$ = number of free electrons per unit volume
$e$ = electron charge …

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Figure — The stem explicitly asks to explain 'with the help of a curve' why weak-electrolyte Λm° cannot be extrapolated

Kohlrausch’s law states that the limiting molar conductivity of an electrolyte is the sum of the independent contributions of its ions. For strong electrolytes, $\Lambda_m$ varies linearly with $\sqrt{c}$ at low concentrations, allowing easy extrapolation to $\Lambda_m^\circ$ . For weak electrolytes, the curve is steep and non-linear near zero concentration, making direct extrapolation unreliable — so we use Kohlrausch’s law instead.

The core idea: Molar conductivity and its limit

Molar conductivity $\Lambda_m$ is the conductivity of a solution containing one mole of electrolyte, placed between electrodes 1 cm apart. As you dilute a solution, $\Lambda_m$ increases because ions are more free to move — they experience less interionic attraction. The value at infinite dilution (zero concentration) is called limiting molar conductivity, $\Lambda_m^\circ$ . This is a fundamental property of the electrolyte, independent of concentration.

The challenge is: how do we find $\Lambda_m^\circ$ for a weak electrolyte like acetic acid? You cannot simply measure it directly because at very low concentrations, weak electrolytes are only partially dissociated, and the conductivity drops sharply.

Kohlrausch’s law of independent migration of ions

\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ

Where $\nu_+$ and $\nu_-$ are the numbers of cations and anions per formula unit, and $\lambda_+^\circ$ , $\lambda_-^\circ$ are their limiting molar conductivities (individual ion contributions). The law says that at infinite dilution, each ion moves independently, and the total conductivity is simply the sum of its parts.

This is powerful because it lets us calculate $\Lambda_m^\circ$ for any electrolyte from known ion values — even for weak electrolytes that cannot be measured directly.

Why extrapolation works for strong electrolytes but fails for weak ones

Strong electrolytes (e.g., NaCl, HCl) At low concentrations, they are fully dissociated. The Debye–Hückel–Onsager theory gives a linear relationship:

\Lambda_m = \Lambda_m^\circ - A\sqrt{c}

where $A$ is a constant. Plot $\Lambda_m$ vs $\sqrt{c}$ , and you get a straight line at low $c$ . Extrapolating to $c = 0$ (i.e., $\sqrt{c} = 0$ ) gives $\Lambda_m^\circ$ directly. This is reliable and precise.

Weak electrolytes (e.g., CH $_3$ COOH)

They are only partially dissociated. As you dilute, the degree of dissociation $\alpha$ increases, so $\Lambda_m$ rises steeply — but not linearly with $\sqrt{c}$ . The curve is concave upward near the origin, bending sharply. Extrapolating a curve that is not linear is guesswork; a tiny error in the slope near zero gives a huge error in the intercept.

The graph below illustrates this:
```
Λ_m
^
|                /  (weak electrolyte)
|              /
|            /
|          /
|        /
|      /
|    /  (strong electrolyte — straight line)
|  /
|/
+-------------------> √c
```
For the weak electrolyte, the curve plunges toward the y-axis so steeply that you cannot reliably draw a tangent to find the intercept. The value of $\Lambda_m^\circ$ is not accessible by direct extrapolation.

Watch out

A common mistake is to assume that the $\Lambda_m$ vs $\sqrt{c}$ plot for a weak electrolyte is linear at low concentrations. It is not — the curvature is severe because $\alpha$ changes rapidly with dilution. Extrapolating a curved line gives a meaningless intercept.

How Kohlrausch’s law solves the problem …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 32 on this concept.

CBSE 2025Set ANNUAL1 mark
$Q. What is the SI unit of molar conductivity?$
›Reveal solutionSolution

Molar conductivity's SI unit is S m² mol⁻¹.

Molar conductivity $\Lambda_m = \dfrac{\kappa}{C}$ , where $\kappa$ (conductivity) has SI unit $S\,m^{-1}$ and concentration $C$ has SI unit $mol\,m^{-3}$ .

$\Lambda_m = \dfrac{S\,m^{-1}}{mol\,m^{-3}} = S\,m^2\,mol^{-1}$

…

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CBSE 2025Set ANNUAL1 mark
$Q. Define the following — Limiting molar conductivity$
›Reveal solutionSolution

Limiting molar conductivity is molar conductivity extrapolated to zero concentration.

Limiting molar conductivity ( $\Lambda_m^0$ or $\Lambda_m^\infty$ ) is the molar conductivity of an electrolyte solution when the concentration approaches zero (i.e. at infinite dilution). At infinite dilution, dissociation of the electrolyte is essentially complete and inter-ionic attractions vanish, so each ion conducts independently and to its maximum extent. For strong electrolytes, $\Lambda_m^0$ is obtained by extrapolating the $\Lambda_m$ …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Equivalent conductances of sodium acetate, sodium chloride and hydrochloric acid at infinite dilution are 224, 38.2, 203 ohm^-1 cm^2 eqv^-1 respectively at 298K. So the (lambda)0 CH3COOH is:$
(a) 288.5 ohm^-1 cm^2 eqv.^-1
(b) 288.8 ohm^-1 cm^2 eqv.^-1
(c) 388.8 ohm^-1 cm^2 eqv.^-1
(d) 59.2 ohm^-1 cm^2 eqv.^-1
›Reveal solutionSolution

λ0(CH3COOH) = λ0(CH3COONa) + λ0(HCl) − λ0(NaCl) = 224 + 203 − 38.2 = 388.8 ohm^-1 cm^2 eqv^-1.

CH3COOH is a weak electrolyte, so its limiting equivalent conductance cannot be found by direct extrapolation. Instead, Kohlrausch's law of independent migration of ions lets us combine the limiting conductances of related strong electrolytes.

We want λ0(CH3COO-) + λ0(H+). Note that:

λ0(CH3COONa) = λ0(CH3COO-) + λ0(Na+) = 224

λ0(HCl) = λ0(H+) + λ0(Cl-) = 203

λ0(NaCl) = λ0(Na+) + λ0(Cl-) = 38.2

…

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CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: Molar conductivity ________ with decrease in concentration.$
›Reveal solutionSolution
Molar conductivity increases as concentration decreases (i.e., on dilution), reaching a maximum limiting value at infinite dilution.

Molar conductivity is given by:

Λm = κ x 1000 / M

As a solution is diluted (concentration M decreases):

For weak electrolytes: the degree of dissociation increases sharply with dilution, so more ions are produced per mole, increasing Λm markedly. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The unit of molar conductivity is$
(a) S cm^-2 mol^-1
(b) S cm^2 mol^-1
(c) S^-1 cm^2 mol^-1
(d) S cm^2 mol
›Reveal solutionSolution

Molar conductivity relates conductivity (S/cm) to concentration (mol/cm^3), giving the composite unit S cm^2 mol^-1.

Molar conductivity is defined as:

$\Lambda_m = \dfrac{\kappa \times 1000}{C}$

where $\kappa$ (specific conductivity) has units S cm^-1 and $C$ (concentration) has units mol L^-1 (mol per 1000 cm^3).

…

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CBSE 2025Set ANNUAL1 markMCQ
$L^{-1}$
(a) 12.9 S cm $^2$ mol $^{-1}$
(b) 1.29 S cm $^2$ mol $^{-1}$
(c) 0.0129 S cm $^2$ mol $^{-1}$
(d) 129 S cm $^2$ mol $^{-1}$
›Reveal solutionSolution

Converting conductivity (per cm) into molar conductivity requires dividing by the molar concentration expressed per cm³, giving a factor of 1000 in the standard formula.

Molar conductivity is related to the specific conductivity (electrolytic conductivity, $\kappa$ ) and molar concentration $C$ (in $mol\,L^{-1}$ ) by:

$\Lambda_m = \frac{\kappa \times 1000}{C}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The formula used to calculate molar conductivity of an electrolyte is _____.$
(a) $\Lambda=\dfrac{1000c}{k}$
(b) $c=\dfrac{1000\Lambda}{k}$
(c) $\Lambda=\dfrac{1000k}{c}$
(d) $k=\dfrac{1000}{\Lambda c}$
›Reveal solutionSolution

$\Lambda_m = \dfrac{1000\,\kappa}{c}$ .

Molar conductivity ( $\Lambda_m$ ) relates to specific conductivity ( $\kappa$ ) and molar concentration $c$ (in $mol\,dm^{-3}$ ) by:

$\Lambda_m = \dfrac{\kappa \times 1000}{c}$

…

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CBSE 2025Set ANNUAL1 mark
$Q. What is the SI unit of molar conductivity? OR Write the relation between specific conductivity and molar conductivity.$
›Reveal solutionSolution

Molar conductivity's SI unit follows from $\Lambda_m = \kappa/C$ : siemens metre-squared per mole.

Molar conductivity is defined as $\Lambda_m = \dfrac{\kappa}{C}$ , where $\kappa$ (specific/electrical conductivity) has SI unit $S\,m^{-1}$ and $C$ (molar concentration) has SI unit $mol\,m^{-3}$ . Dividing, the SI unit of $\Lambda_m$ works out to

$\frac{S\,m^{-1}}{mol\,m^{-3}} = S\,m^2\,mol^{-1}.$

(In practical lab work, where $\kappa$ is often expressed in $S\,cm^{-1}$ and $C$ in $mol\,L^{-1}$ , the commonly used relation is $\Lambda_m = \dfrac{1000\,\kappa}{C}$ , giving the c.g.s.-style unit $S\,cm^2\,mol^{-1}$ .)

…

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CBSE 2024Set A11 markMCQ
$Q. When the concentration of electrolytic solution approaches zero, the resulting molar conductivity is known as ;$
(a) specific conductance
(b) resistivity
(c) conductivity
(d) limiting molar conductivity
›Reveal solutionSolution

Molar conductivity at zero concentration (infinite dilution) is called the limiting molar conductivity — option (d).

Molar conductivity $\Lambda_m$ increases as an electrolytic solution is diluted, because more of the electrolyte is present as free, effectively conducting ions. As the concentration approaches zero (infinite dilution), $\Lambda_m$ reaches a limiting maximum value denoted $\Lambda_m^\circ$ , the **limiting molar co …
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CBSE 2024Set B1 mark
$Q. Answer in one word/sentence: Write the unit of Equivalence conductivity.$
›Reveal solutionSolution

Equivalent conductivity is conductivity per gram-equivalent of electrolyte per unit volume, giving units of S cm^2 eq^-1.

Equivalent conductivity, $\Lambda_{eq}$ , is defined as $\Lambda_{eq} = \kappa \times V$ , where $\kappa$ (specific conductivity) has units of S cm^-1 (ohm^-1 cm^-1) and V is the volume in cm^3 containing one gram-equivalent of …

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CBSE 2024Set ANNUAL1 mark
$2.5 \times 10^{-2}$
›Reveal solutionSolution

Adding the limiting ionic molar conductivities gives Λm°, and dividing the measured Λm by it gives the degree of dissociation.

For a weak electrolyte, the degree of dissociation is given by:

$\alpha = \frac{\Lambda_m}{\Lambda_m°}$

Limiting molar conductivity of methanoic acid (using Kohlrausch's law of independent migration of ions):

$\Lambda_m° = \lambda°_{H^+} + \lambda°_{HCOO^-} = 349.6 + 54.6 = 404.2\ S\,cm^2\,mol^{-1}$ …

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CBSE 2024Set ANNUAL1 mark
$Q. Write SI unit of molar conductivity.$
›Reveal solutionSolution

SI unit of molar conductivity is $S\,m^2\,mol^{-1}$ .

Molar conductivity $\Lambda_m$ is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte, measured between electrodes 1 m apart.

…

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Q.State Kohlrausch's law of independent migration of ions. With the help of a curve, explain why it is not easy to determine Λm∘\Lambda_m^\circΛm∘​ for weak electrolytes by extrapolating the concentration −-− molar conductivity curve, as it is for strong electrolytes.

The core idea: Molar conductivity and its limit

Kohlrausch’s law of independent migration of ions

Why extrapolation works for strong electrolytes but fails for weak ones

How Kohlrausch’s law solves the problem …

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Q.State Kohlrausch's law of independent migration of ions. With the help of a curve, explain why it is not easy to determine $\Lambda_m^\circ$ for weak electrolytes by extrapolating the concentration $-$ molar conductivity curve, as it is for strong electrolytes.

Unlock everything free for 14 days