(a) (i) The resistance of 0·05 M $CH_3COOH$ solution is found to be 100 ohm. If the cell constant is 0·0354 cm$^{-1}$, calculate the molar conductivity of the acetic acid solution. (ii) Write Faraday's first law of electrolysis. How many Faraday is required for the reduction of 1 mol of $MnO_4^-$ to $Mn^{2+}$ ? OR (b) (i) The conductivity of 0·0025 mol L$^{-1}$ acetic acid is $5·25 \times 10^{-5}$ S cm$^{-1}$. Calculate its degree of dissociation if $\Lambda^\circ_m$ for acetic acid is 390 S cm$^2$ mol$^{-1}$. (ii) Write anode, cathode and overall reaction of lead storage battery.

(a) (i) The resistance of 0·05 M CH₃COOH solution is found to be 100 ohm. If the cell constant is 0·0354 cm⁻¹,

Question

Q. (a)

(i) The resistance of 0·05 M

CH_3COOH

solution is found to be 100 ohm. If the cell constant is 0·0354 cm

^{-1}

, calculate the molar conductivity of the acetic acid solution.

(ii) Write Faraday's first law of electrolysis. How many Faraday is required for the reduction of 1 mol of

MnO_4^-

Mn^{2+}

(OR)

(b)

(i) The conductivity of 0·0025 mol L

^{-1}

acetic acid is

5·25 \times 10^{-5}

S cm

^{-1}

. Calculate its degree of dissociation if

\Lambda^\circ_m

for acetic acid is 390 S cm

^2

mol

^{-1}

(ii) Write anode, cathode and overall reaction of lead storage battery.

CBSECBSE Class XII Board 2024Subjective· 5mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Conductance And 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 — Conductivity of Electrolytes

Conductivity of Electrolytes – From Intuition to Precision

Think of a copper wire. You know it conducts electricity because electrons flow through it. Now imagine dipping two metal plates into a beaker of salt water and connecting them to a battery. The bulb glows. The salt water is conducting — but not with electrons. Something else is carrying the charge.

That something is ions. When an electrolyte (like NaCl, HCl, or NaOH) dissolves in water, it splits into positive and negative ions. These ions are free to move. When you apply a voltage, positive ions drift toward the negative electrode, negative ions toward the positive electrode. That directed motion of charged particles is an electric current. That is the core idea: electrolytic conductivity is the ability of a solution to carry current via the movement of its ions.

Note

The more ions present, and the faster they can move, the higher the conductivity. But it is not that simple — concentration changes both the number of ions and how they interact with each other.

The Precise Definition

Conductivity (symbol $\kappa$ , units S m $^{-1}$ or S cm $^{-1}$ ) is the reciprocal of resistivity. For a solution placed between two parallel electrodes of area $A$ and separation $l$ , the resistance $R$ is:

R = \frac{1}{\kappa} \cdot \frac{l}{A}

So $\kappa$ is the conductance of a 1 m × 1 m × 1 m cube of the solution. It depends on:

Number of ions per unit volume (concentration)
Charge on each ion
How fast the ions move (their mobility)

But here is the catch: as you dilute a solution, $\kappa$ does not simply drop proportionally. Why? Because dilution changes both the number of ions and the degree of dissociation (for weak electrolytes). To compare the conducting power of different electrolytes fairly, we need a quantity that normalises for concentration.

Molar Conductivity – The Fair Comparison

Molar conductivity ( $\Lambda_m$ , units S m $^2$ mol $^{-1}$ ) is defined as:

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

where $c$ is the concentration in mol m $^{-3}$ . It tells you: if I had exactly one mole of electrolyte dissolved in a solution, what would the conductivity of that entire solution be?

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

For a strong electrolyte like KCl, $\Lambda_m$ decreases slowly as concentration increases. Why? Because at higher concentrations, ions are closer together — they feel each other's electric fields, slow each other down (ion-ion interactions). At infinite dilution ( $c \to 0$ ), ions are so far apart they move independently, and $\Lambda_m$ reaches a maximum value called $\Lambda_m^\infty$ (limiting molar conductivity).

For a weak electrolyte like acetic acid, $\Lambda_m$ rises sharply on dilution. That is because dilution increases the degree of dissociation — more ions are formed from the same number of molecules. At infinite dilution, all molecules are dissociated, and $\Lambda_m^\infty$ is the sum of the individual ion contributions (Kohlrausch's law). …

Start your 14-day free trial to unlock the full solution →

Part (a): $\Lambda_m=7.08\ \text{S cm}^2\text{mol}^{-1}$ ; Faraday's first law ( $m\propto Q$ ) and 5 F reduce 1 mol $MnO_4^-\to Mn^{2+}$ . Part (b): $\alpha=0.0538$ (5.38%); lead-acid battery discharge overall $Pb+PbO_2+2H_2SO_4\to2PbSO_4+2H_2O$ .

Key relations: $\kappa=\dfrac{\text{cell constant}}{R}$ , $\Lambda_m=\dfrac{\kappa\times1000}{c}$ , $\alpha=\dfrac{\Lambda_m}{\Lambda^\circ_m}$ .

Part (a)

(i) From resistance and cell constant, the conductivity is

\kappa=\frac{G^*}{R}=\frac{0.0354\ \text{cm}^{-1}}{100\ \Omega}=3.54\times10^{-4}\ \text{S cm}^{-1}.

Then with $M=0.05\ \text{mol L}^{-1}$ ,

$\Lambda_m=\frac{\kappa\times1000}{M}=\frac{3.54\times10^{-4}\times1000}{0.05}=\frac{0.354}{0.05}=7.08\ \text{S cm}^2\text{mol}^{-1}.$ …

Start your 14-day free trial to unlock the full solution →

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 42 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, …

Start your 14-day free trial to see the full solution →
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-

…

Start your 14-day free trial to see the full solution →
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 $\dfrac{cm}{cm^2} = cm^{-1}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Conductivity of electrolytic solutions ______ with increase of temperature (fill in the blank).$
›Reveal solutionSolution

Raising the temperature increases ionic mobility (lower viscosity, faster ion movement) and, for weak electrolytes, increases the degree of dissociation, so conductivity goes up.

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Conductivity of a solution increases with dilution.$
›Reveal solutionSolution

False - conductivity decreases on dilution (molar conductivity increases).

Conductivity (specific conductance, kappa) is the conductance of ions present in 1 cm3 of solution. On dilution the total number of ions per unit volume decreases, so kappa decreases.

…

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Why does the conductivity of a solution decrease with dilution?$
›Reveal solutionSolution

Conductivity (κ) is conductance per unit volume; diluting spreads the same ions over more volume, so ion density falls.

Conductivity ( $\kappa$ ) is a measure of the conducting power per unit volume of the solution, i.e. it depends on the number of ions present per unit volume.

On dilution, the total volume of the solution increases while the total number of ions (for a fixed amount of electrolyte) essentially stays the same (or increases only slightly for weak electrolytes due to greater dissociation) — so the concentration of ions per unit volume decreases. Since $\kappa$ is directly proportional to ion concentration, $\kappa$ decreases with d …

Start your 14-day free trial to see the full solution →
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:

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. A weak electrolyte:$
(a) Does not dissociate into ions.
(b) Dissociate completely into ions.
(c) Dissociate into ions but not completely.
(d) None of these.
›Reveal solutionSolution
A weak electrolyte partially ionizes in solution (degree of dissociation $\alpha \ll 1$ ), unlike a strong electrolyte which ionizes almost completely.

Electrolytes are substances whose aqueous solutions conduct electricity because they furnish ions. They are classified by how completely they ionize:

Strong electrolytes (e.g. NaCl, HCl, KOH) dissociate almost completely (~100%) into ions in solution. …
Start your 14-day free trial to see the full solution →

Q.(a)

Part (a)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days