(a) Conductivity of $2 \times 10^{-3}$ M methanoic acid is $8 \times 10^{-5}$ S cm$^{-1}$. Calculate its molar conductivity and degree of dissociation if $\Lambda^\circ_m$ for methanoic acid is 404 S cm$^2$ mol$^{-1}$. (3 + 2) (b) Calculate the $\Delta_rG^\circ$ and log $K_c$ for the given reaction at 298 K : $Ni(s) + 2Ag^+(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)$ Given : $E^\circ_{Ni^{2+}/Ni} = -0.25$ V, $E^\circ_{Ag^+/Ag} = +0.80$ V $1F = 96500$ C mol$^{-1}$.

(a) Conductivity of 2 × 10⁻³ M methanoic acid is 8 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity and degree

Question

2 \times 10^{-3}

(b) Calculate the

\Delta_rG^\circ

and log

K_c

for the given reaction at 298 K :

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

Given :

E^\circ_{Ni^{2+}/Ni} = -0.25

E^\circ_{Ag^+/Ag} = +0.80

1F = 96500

C mol

^{-1}

CBSECBSE Class XII Board 2023Subjective· 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 →

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 …

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

For part (a), we use the relation between conductivity and molar conductivity to find $\Lambda_m$ , then compare it with $\Lambda_m^\circ$ to get the degree of dissociation $\alpha$ . For part (b), we compute the standard cell potential, then use $\Delta_r G^\circ = -nFE^\circ_{cell}$ and $\Delta_r G^\circ = -2.303RT \log K_c$ to find $\Delta_r G^\circ$ and $\log K_c$ .

Part (a) — Methanoic acid conductivity

Concept first: Molar conductivity $\Lambda_m$ tells us how well a solution conducts electricity per mole of electrolyte. It is directly linked to the measured conductivity $\kappa$ (kappa) by:

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

where $C$ is concentration in mol L $^{-1}$ . The factor 1000 converts cm $^3$ to L, because conductivity is in S cm $^{-1}$ and we want $\Lambda_m$ in S cm $^2$ mol $^{-1}$ .

For a weak acid like methanoic acid, the degree of dissociation $\alpha$ is the fraction of molecules that have ionised. At a given concentration, the measured molar conductivity $\Lambda_m$ is less than the limiting molar conductivity $\Lambda_m^\circ$ (at infinite dilution) because not all molecules are dissociated. For weak electrolytes, Kohlrausch’s law gives:

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

Watch out

This relation $\alpha = \Lambda_m / \Lambda_m^\circ$ holds strictly only for weak electrolytes. For strong electrolytes, $\Lambda_m$ changes with concentration due to interionic attractions, not just dissociation — so this formula is not used there.

Step-by-step:

Given data:

$\kappa = 8 \times 10^{-5}$ S cm $^{-1}$

$C = 2 \times 10^{-3}$ M = $2 \times 10^{-3}$ mol L $^{-1}$

$\Lambda_m^\circ = 404$ S cm $^2$ mol $^{-1}$
Calculate molar conductivity $\Lambda_m$ :

\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{8 \times 10^{-5} \times 1000}{2 \times 10^{-3}}

Simplify stepwise:

$8 \times 10^{-5} \times 1000 = 8 \times 10^{-2}$

Divide by $2 \times 10^{-3}$ :

\Lambda_m = \frac{8 \times 10^{-2}}{2 \times 10^{-3}} = 4 \times 10^{1} = 40 \text{ S cm}^2 \text{ mol}^{-1}

Calculate degree of dissociation $\alpha$ :

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{40}{404}

Simplify:

\alpha = 0.0990 \quad (\text{approximately } 0.099)

In percentage terms, about 9.9% dissociation.

Tip

Notice that $\Lambda_m$ (40) is much smaller than $\Lambda_m^\circ$ (404) — this is typical for a weak acid. The low $\alpha$ (~0.1) confirms methanoic acid is weakly dissociated at this concentration.

Part (b) — $\Delta_r G^\circ$ and $\log K_c$ for the cell reaction

Concept first: For any electrochemical cell, the standard Gibbs free energy change is related to the standard cell potential by:

\Delta_r G^\circ = -nFE^\circ_{cell}

where $n$ is the number of moles of electrons transferred in the balanced reaction, $F$ is Faraday’s constant (96500 C mol $^{-1}$ ), and $E^\circ_{cell}$ is the standard cell potential.

The equilibrium constant $K_c$ is related to $\Delta_r G^\circ$ via:

\Delta_r G^\circ = -2.303 RT \log K_c

At 298 K, $2.303 RT = 2.303 \times 8.314 \times 298 \approx 5710$ J mol $^{-1}$ (or 5.71 kJ mol $^{-1}$ ). Often we use the form:

\log K_c = \frac{n E^\circ_{cell}}{0.0591} \quad \text{(at 298 K, with } E^\circ \text{ in volts)}

because $\frac{2.303 RT}{F} = 0.0591$ V at 298 K.

Step-by-step:

Identify half-reactions and $n$ : Anode (oxidation): $\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^-$ Cathode (reduction): $2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag}(s)$ Overall: $\text{Ni}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$ Number of electrons transferred: $n = 2$ . …

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 27 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 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 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 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 2024Set D1 markMCQ
$Q. Which of the following has the highest molar electrical conductance in aqueous solution?$
(a) [Pt(NH3)6]Cl4
(b) [Pt(NH3)5Cl]Cl3
(c) [Pt(NH3)4Cl2]Cl2
(d) [Pt(NH3)3Cl3]Cl
›Reveal solutionSolution
Molar conductance rises with the number of ions produced on dissociation. [Pt(NH3)6]Cl4 gives 5 ions, the most of the options, so it conducts best.

Count the ions each complex furnishes in water (only the counter-ions outside the coordination sphere ionise):

[Pt(NH3)6]Cl4 -> [Pt(NH3)6]4+ + 4 Cl- => 5 ions

[Pt(NH3)5Cl]Cl3 -> [Pt(NH3)5Cl]3+ + 3 Cl- => 4 ions

[Pt(NH3)4Cl2]Cl2 -> [Pt(NH3)4Cl2]2+ + 2 Cl- => 3 ions …
Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$Q. The cell constant of a conductivity cell is$
(a) l/A
(b) A/l
(c) l.A
(d) R/A
›Reveal solutionSolution

Cell constant = l/A (distance between electrodes ÷ electrode area), unit cm^-1.

Conductance G of a solution in a conductivity cell is G = kappa (A/l), where kappa is conductivity, A is the electrode area and l is the distance between the electrodes.

Rearranging, kappa = G (l/A). The geometric factor (l/A) is called the CELL CONSTANT because it depends only on the fixed geometry of the cell.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set B1 markMCQ
$Q. The unit of cell constant is$
(a) ohm cm
(b) cm^-1
(c) cm
(d) ohm^-1 cm^-1
›Reveal solutionSolution

Cell constant G* = l/A (distance between electrodes divided by their area), so its unit works out to cm^-1.

For a conductivity cell, the cell constant is defined as:

$G^{*} = \frac{l}{A}$

where l is the distance between the two electrodes (in cm) and A is the area of cross-section of the electrodes (in cm^2).

Units: cm / cm^2 = cm^-1.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. SI unit of conductivity is -$
(a) S
(b) Ω
(c) S cm^-1
(d) S m^-1
›Reveal solutionSolution

Conductivity kappa = conductance x (l/A); in SI units this comes out as siemens per metre (S m^-1).

Conductance G has the unit siemens (S), where S = ohm^-1 = A/V.

Conductivity kappa is defined as kappa = G x (l/A), where l is the length between electrodes (in metres) and A is the cross-sectional area (in square metres).

So the unit of kappa = S x (m/m^2) = S m^-1.

Option (a) S is just the unit of conductance, not conductivity. …

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

Q.(a) Conductivity of 2×10−32 \times 10^{-3}2×10−3 M methanoic acid is 8×10−58 \times 10^{-5}8×10−5 S cm−1^{-1}−1. Calculate its molar conductivity and degree of dissociation if Λm∘\Lambda^\circ_mΛm∘​ for methanoic acid is 404 S cm2^22 mol−1^{-1}−1. (3 + 2)

Part (a) — Methanoic acid conductivity

Part (b) — ΔrG∘\Delta_r G^\circΔr​G∘ and log⁡Kc\log K_clogKc​ for the cell reaction

🎓Unlock everything free for 14 days

Q.(a) Conductivity of $2 \times 10^{-3}$ M methanoic acid is $8 \times 10^{-5}$ S cm $^{-1}$ . Calculate its molar conductivity and degree of dissociation if $\Lambda^\circ_m$ for methanoic acid is 404 S cm $^2$ mol $^{-1}$ . (3 + 2)

Part (b) — $\Delta_r G^\circ$ and $\log K_c$ for the cell reaction

Unlock everything free for 14 days