(A) (a) The conductivity of 0.1 mol L⁻¹ solution of NaCl is 1.06 × 10⁻² S cm⁻¹. Calculate its molar conductivi

Question

Q. (A)

(a) The conductivity of 0.1 mol L

^{-1}

solution of NaCl is

1.06 \times 10^{-2}

S cm

^{-1}

. Calculate its molar conductivity and degree of dissociation.

\lambda^o_{Na^+} = 50.1

S cm

^2

mol

^{-1}

\lambda^o_{Cl^-} = 76.5

S cm

^2

mol

^{-1}

(b)

(i) Following cell reaction occurs in a galvanic cell :

2Ag^+(aq) + Zn(s) \rightarrow 2Ag(s) + Zn^{2+}(aq)

E^o_{(cell)} = +1.56

V Predict the direction of flow of current.

(ii) Differentiate between a primary battery and a secondary battery.

(OR)

(B)

(a) Resistance of a conductivity cell filled with 0.1 M KCl solution is 100

\Omega

. If the resistance of the same cell when filled with 0.01 mol L

^{-1}

KCl solution is 300

\Omega

, calculate the conductivity and molar conductivity of 0.01 mol L

^{-1}

KCl solution. The conductivity of 0.1 M KCl solution is

1.29 \times 10^{-2}

S cm

^{-1}

(b)

(i) Write any two advantages of

H_2 - O_2

fuel cell.

(ii) Why does the cell potential of mercury cell remain constant throughout the life ?

CBSECBSE Class XII Board 2026Subjective· 5mImportance★★★★★

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Part (a)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 …

Part (b)Concept understanding — Conductance And Conductivity

From Resistance to Conductance: Flipping the Idea

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}

\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

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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Part (a): NaCl $\Lambda_m = 106$ S cm $^2$ mol $^{-1}$ , $\alpha = 0.837$ ; current flows Ag→Zn externally; primary = non-rechargeable, secondary = rechargeable.

Part (b): 0.01 M KCl $\kappa = 4.3\times10^{-3}$ S cm $^{-1}$ , $\Lambda_m = 430$ S cm $^2$ mol $^{-1}$ ; fuel cell is efficient/clean; mercury cell holds constant voltage.

Part (a)

(a) NaCl molar conductivity and degree of dissociation

\Lambda_m = \frac{\kappa\times1000}{C} = \frac{1.06\times10^{-2}\times1000}{0.1} = 106\ \text{S cm}^2\ \text{mol}^{-1}

\Lambda_m^\circ = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} = 50.1 + 76.5 = 126.6\ \text{S cm}^2\ \text{mol}^{-1}

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{106}{126.6} = 0.837

(For a strong electrolyte this is an apparent degree of dissociation.)

(b)(i) Direction of current flow

With $E^\circ_{cell} = +1.56$ V the reaction $2Ag^+ + Zn \rightarrow 2Ag + Zn^{2+}$ is spontaneous: Zn is oxidised (anode, −), $Ag^+$ reduced (cathode, +). Electrons flow Zn → Ag through the wire, so conventional current flows from the silver electrode to the zinc electrode in the external circuit.

(b)(ii) Primary vs secondary battery

Primary	Secondary
Not rechargeable, irreversible reaction	Rechargeable, reversible reaction

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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 59 on this concept.

CBSE 2026Set A1 markMCQ
$Q. On increasing dilution, the specific conductance of an electrolyte$
(a) increases
(b) decreases
(c) remains constant
(d) none of these
›Reveal solutionSolution

Specific conductance (conductance per unit volume) falls on dilution because the number of current-carrying ions per unit volume decreases.

Specific conductance (κ) is the conductance of a solution held between electrodes 1 cm apart with 1 cm² area, i.e. conductance of unit volume. On dilution the number of ions per unit volume decreases, …

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CBSE 2026Set A1 markMCQ
$Q. The number of ions in aqueous solution of [Co(NH3)5Cl]Cl2 is$
(a) 3
(b) 4
(c) 2
(d) 6
›Reveal solutionSolution

Only the ions outside the coordination sphere are free; [Co(NH3)5Cl]Cl2 gives one complex cation plus two chloride ions = 3 ions.

In a coordination compound, only the counter ions outside the square brackets dissociate in water; the ligands inside the coordination sphere stay bound to the metal. Here one Cl and five NH3 are coordinated to cobalt, and two Cl are counter ions:

[Co(NH3)5Cl]Cl2 -> [Co(NH3)5Cl]2+ + 2 Cl-

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The unit of cell constant is:$
(a) Ohm^-1 cm^2
(b) cm^-1
(c) Ohm^-1 cm^-1
(d) Ohm^-1 cm^2/ g eq
›Reveal solutionSolution

Cell constant $G^{*}=l/A$ has the unit of reciprocal length, i.e. cm^-1.

The cell constant of a conductivity cell is defined as the ratio of the distance between the two electrodes (l) to the area of cross-section of the electrodes (A): $G^{*} = \dfrac{l}{A}$ . Since l has units of cm and A has units of cm^2, the cell constant has units of $\dfrac{cm}{cm^2} = cm^{-1}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The unit of specific conductivity is:$
(a) ohm⁻¹
(b) ohm⁻¹ cm⁻¹
(c) ohm cm
(d) ohm cm⁻¹
›Reveal solutionSolution

Specific conductance ( $\kappa$ ) is measured in ohm⁻¹ cm⁻¹ (S cm⁻¹).

Specific conductivity ( $\kappa$ ), also called conductivity, is the conductance of a 1 cm cube of a solution of an electrolyte. Conductance ( $G$ ) is the reciprocal of resistance and is measured in ohm⁻¹ (siemens, S). Since $\kappa = G \times (l/A)$ , where $l/A$ (t …

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CBSE 2025Set 56/5/11 markMCQ
$Q. Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : For measuring resistance of an ionic solution an AC source is used. Reason (R) : Concentration of ionic solution will change if DC source is used.$
›Reveal solutionSolution

AC is used to measure ionic solution resistance because DC causes electrolysis, which changes the solution's composition and hence its conductance; both statements are true and the reason correctly explains the assertion.

When we measure the resistance (or conductance) of an ionic solution, we're essentially probing how easily ions can carry current through the liquid. The choice between AC and DC isn't arbitrary—it stems from what happens at the electrode-solution interface.

Why DC causes problems

In an ionic solution, current flows via the movement of ions: cations migrate toward the cathode, anions toward the anode. With a DC source, these ions don't just move—they undergo redox reactions at the electrodes. For instance, in a $\text{NaCl}$ solution, $\text{Cl}^-$ ions get oxidized at the anode ( $2\text{Cl}^- \to \text{Cl}_2 + 2e^-$ ) and $\text{H}^+$ from water gets reduced at the cathode ( $2\text{H}^+ + 2e^- \to \text{H}_2$ ). This is electrolysis.

The consequence? The concentration of ions in the solution changes continuously. As ions are consumed or new species are produced, the conductance of the solution drifts. You're no longer measuring the property of the original solution—you're measuring a changing system. The reading becomes unreliable and time-dependent.

Why AC solves this

An alternating current reverses direction many times per second (typically at 1000 Hz or so in conductivity bridges). In one half-cycle, a tiny bit of electrolysis might begin, but in the next half-cycle the current reverses and the reaction is essentially undone. The net chemical change over many cycles is negligible. The solution composition remains stable, and the resistance measurement reflects the true, steady-state property of the ionic solution. …
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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 D1 markMCQ
$Q. The unit of specific conductance is$
(a) ohm cm^-1
(b) ohm cm^-2
(c) ohm^-1 cm^-1
(d) ohm^-1 cm^-2
›Reveal solutionSolution

Specific conductance = 1/(specific resistance), so its unit is ohm^-1 cm^-1 (S cm^-1).

Specific conductance (conductivity), kappa, is the reciprocal of specific resistance (resistivity), rho:

kappa = 1/rho

Specific resistance has the unit ohm cm, so its reciprocal has the unit:

…

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CBSE 2025Set A1 mark
$Q. Write the value of conductivity of superconductor.$
›Reveal solutionSolution

Since conductivity is the reciprocal of resistivity, and a superconductor's resistivity drops to exactly zero, its conductivity becomes infinite.

Certain materials, when cooled below a characteristic critical temperature, lose all electrical resistance completely — this state is called superconductivity, and such materials are superconductors. Electrical conductivity ( $\kappa$ ) and resistivity ( $\rho$ ) are reciprocals of each other: $\kappa = 1/\rho$ . Because a superconductor's res …

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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 markMCQ
$Q. SI unit of resistivity (specific resistance) is -$
(a) Ω
(b) Ω^-1
(c) Ωm
(d) Ωm^-1
›Reveal solutionSolution

Resistivity (specific resistance) has SI unit ohm-metre (Ωm).

Resistance of a conductor is related to its resistivity by:

R = rho x (l/A)

where l is length (m) and A is cross-sectional area (m^2). Rearranging:

rho = R x A / l

Units: rho = (ohm) x (m^2) / (m) = ohm x m = Ωm

…

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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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Q.(A)

Part (a)

(a) NaCl molar conductivity and degree of dissociation

(b)(i) Direction of current flow

(b)(ii) Primary vs secondary battery

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