Match the terms given in Column I with the units given in Column II. Column I: (i) Lambdam (ii) ECell (iii) ka

NCERT Exemplar · Q53

Q. Match the terms given in Column I with the units given in Column II. Column I:

(i)

\Lambda_m

(ii)

E_{Cell}

(iii)

\kappa

(iv)

G^*

Column II:

(a)

S\ cm^{-1}

(b)

m^{-1}

(c)

S\ cm^2\ mol^{-1}

(d)

V

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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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This is a matching problem linking four electrochemical quantities to their correct SI or practical units. The key is to recall the definitions: molar conductivity $\Lambda_m$ is conductivity per mole, so its unit is $S\ cm^2\ mol^{-1}$ ; cell potential $E_{Cell}$ is voltage, so its unit is $V$ ; conductivity $\kappa$ is the reciprocal of resistivity, so its unit is $S\ cm^{-1}$ ; and the cell constant $G^*$ is the ratio of length to area, so its unit is $m^{-1}$ . The correct matches are: (i)→(c), (ii)→(d), (iii)→(a), (iv)→(b).

The question asks you to match four fundamental electrochemical quantities with their units. This is not about memorising a table — it is about understanding what each quantity physically represents. Once you know the definition, the unit follows naturally.

Let us go through each one.

1. Molar conductivity $\Lambda_m$

Molar conductivity is defined as the conductivity of a solution divided by the molar concentration of the electrolyte:

\Lambda_m = \frac{\kappa}{c}

Here $\kappa$ has units of $S\ cm^{-1}$ (siemens per centimetre) and $c$ has units of $mol\ cm^{-3}$ (or more commonly $mol\ L^{-1}$ , but we must be careful with volume units). In practice, $\kappa$ is measured in $S\ cm^{-1}$ and concentration in $mol\ cm^{-3}$ , so:

\text{Unit of } \Lambda_m = \frac{S\ cm^{-1}}{mol\ cm^{-3}} = S\ cm^2\ mol^{-1}

That matches option (c).

Tip

A common shortcut: molar conductivity always has "per mole" in the denominator and "area" in the numerator — think of it as conductivity per unit concentration, which gives $S\ cm^2\ mol^{-1}$ .

2. Cell potential $E_{Cell}$

The electromotive force (emf) of a cell is simply the potential difference between its two electrodes. The SI unit of electric potential is the volt ( $V$ ). So $E_{Cell}$ is measured in volts. That matches option (d).

3. Conductivity $\kappa$

Conductivity is the reciprocal of resistivity ( $\rho$ ). Resistivity has units of $\Omega\ cm$ (ohm-centimetre), so:

\kappa = \frac{1}{\rho} \quad \Rightarrow \quad \text{Unit of } \kappa = \frac{1}{\Omega\ cm} = S\ cm^{-1}

Here $S$ (siemens) is the unit of conductance, equal to $\Omega^{-1}$ . So $\kappa$ has units $S\ cm^{-1}$ , matching option (a). …

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Method: Unit Analysis by Dimensional Matching

This method works by recalling the defining formula for each term, then deriving its unit from the SI/base units of the quantities in the formula.

Step 1: Recall the definition of each term

Term	Definition / Formula	Key quantities involved
$\Lambda_m$ (Molar conductivity)	$\Lambda_m = \frac{\kappa}{c}$	$\kappa$ = conductivity, $c$ = concentration (mol/cm³)
$E_{Cell}$ (Cell potential)	$E = -\frac{\Delta G}{nF}$	Voltage = J/C = V
$\kappa$ (Conductivity)	$\kappa = \frac{1}{\rho}$ (reciprocal of resistivity)	Unit: S/cm or S/m
$G^*$ (Cell constant)	$G^* = \frac{l}{A}$	$l$ = length, $A$ = area → unit: cm⁻¹ or m⁻¹

Step 2: Derive the unit for each

$\Lambda_m$ :

$\kappa$ has unit $S\,cm^{-1}$ , $c$ has unit $mol\,cm^{-3}$

$\Rightarrow \Lambda_m = \frac{S\,cm^{-1}}{mol\,cm^{-3}} = S\,cm^2\,mol^{-1}$

→ matches (c)
$E_{Cell}$ :

Directly measured in volts

→ matches (d)
$\kappa$ :

Conductivity unit is siemens per centimetre …

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Here are the common mistakes students make when matching these electrochemical quantities with their units, and how to avoid each.

Mistake 1: Confusing $\Lambda_m$ (Molar Conductivity) with $\kappa$ (Conductivity)

The error:

Students often swap the units of molar conductivity ( $\Lambda_m$ ) and specific conductivity ( $\kappa$ ). They match $\Lambda_m$ with $S\ cm^{-1}$ (conductivity) and $\kappa$ with $S\ cm^2\ mol^{-1}$ (molar conductivity).

Why it happens:

Both terms sound similar and both involve "conductivity." The formula $\Lambda_m = \frac{\kappa}{C}$ makes it easy to mix them up.

How to avoid:

Remember the definition:

Molar conductivity ( $\Lambda_m$ ) is the conductivity of a solution containing 1 mole of electrolyte. Its unit must involve per mole → $S\ cm^2\ mol^{-1}$ .
Conductivity ( $\kappa$ ) is the conductance of a 1 cm cube of solution → unit is $S\ cm^{-1}$ .

Trick: The "molar" in $\Lambda_m$ means "per mole" → look for $mol^{-1}$ in the unit.

Mistake 2: Forgetting that $E_{Cell}$ is measured in Volts

The error:

Some students match $E_{Cell}$ with $S\ cm^{-1}$ or $m^{-1}$ , thinking it is a conductivity-related term.

Why it happens:

$E_{Cell}$ appears alongside conductivity terms, so students assume it must also be a conductivity unit.

How to avoid:

Recall that $E$ stands for electromotive force (emf) or cell potential. The unit of potential difference is always the volt (V). No other unit in Column II has $V$ .

Key fact: $E_{Cell} = E^\circ_{cathode} - E^\circ_{anode}$ → result is in volts.

Mistake 3: Confusing $G^*$ (Cell Constant) with Conductivity

The error:

Students match $G^*$ with $S\ cm^{-1}$ or $S\ cm^2\ mol^{-1}$ , thinking it is a type of conductance.

Why it happens:

The symbol $G$ is often used for conductance, so $G^*$ looks like it should have a conductance unit.

How to avoid:

Know the definition:

$G^*$ (cell constant) = $\frac{l}{A}$ , where $l$ is length and $A$ is area. Its unit is $cm^{-1}$ or $m^{-1}$ . It is not a conductance — it is a geometric factor.

Trick: The star (*) means it is a constant for the cell, not a measured quantity. Units are inverse length.

Mistake 4: Mixing up $cm^{-1}$ and $m^{-1}$

The error:

Students match $G^*$ with $S\ cm^{-1}$ (conductivity) instead of $m^{-1}$ . …

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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.Match the terms given in Column I with the units given in Column II.
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