In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

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Q. In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

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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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Specific conductivity (κ) of an electrolyte solution decreases when water is added, because dilution reduces the number of charge carriers per unit volume, even though the ions themselves become more mobile.

Why this happens — the core idea

Specific conductivity (κ) measures how well a fixed volume of solution conducts electricity. It depends on two things: how many ions are packed into that volume, and how fast each ion can move. When you add water, you're literally spreading the same number of ions over a larger volume. The concentration drops, so the ion density drops — and κ falls.

But here's the twist: as the solution gets more dilute, ions have more room to move and less interionic attraction to slow them down. Their individual mobility actually increases. So why doesn't κ go up? Because the drop in the number of ions per unit volume is far more dramatic than the gain in mobility. The net effect is a decrease.

Watch out

A common mistake is to confuse specific conductivity (κ) with molar conductivity (Λₘ). Molar conductivity increases on dilution because it accounts for the number of ions per mole — but specific conductivity, which is a bulk property of the solution, always decreases.

Step-by-step reasoning

Define specific conductivity (κ) It is the conductance of a 1 cm cube of solution, measured in S cm⁻¹. It depends directly on the concentration of ions ( $c$ ) and their mobility ( $u$ ):

\kappa = \sum_i c_i \, |z_i| \, F \, u_i

where $F$ is Faraday's constant and $z_i$ is the charge number. The key point: κ is proportional to concentration of ions.

What happens when water is added?

Adding water dilutes the solution. The number of moles of electrolyte stays the same, but the volume increases. So the concentration $c$ (mol L⁻¹) decreases.
Effect on ion density

Fewer ions per unit volume means fewer charge carriers available to conduct electricity through a fixed cross-section. This directly reduces κ.
Effect on ion mobility

In a concentrated solution, ions are crowded and experience strong interionic attractions, which slows them down. On dilution, these attractions weaken, and ions move more freely. So mobility $u$ increases slightly.
Which effect dominates? …

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Here are the most common mistakes students make when asked about the effect of adding water on specific conductivity (κ), along with the correct reasoning and how to avoid each.

Mistake 1: Confusing Specific Conductivity with Molar Conductivity

The error:

Students often say “conductivity increases because dilution increases dissociation.” This is wrong for specific conductivity (κ), but correct for molar conductivity (Λₘ).

Why it’s wrong:

Specific conductivity (κ) is the conductance of a 1 cm × 1 cm column of solution.
Adding water dilutes the number of ions per unit volume.
Even though each ion moves faster (due to less interionic attraction), the drop in ion concentration dominates → κ decreases.

How to avoid:

Memorise the key distinction:
- κ (specific) → decreases with dilution.
- Λₘ (molar) → increases with dilution (for weak electrolytes, sharply).
Write the definition before answering:

\kappa = \frac{1}{\rho} \cdot \frac{1}{R} \quad \text{(conductance per unit volume)}

Dilution → fewer charge carriers per cm³ → κ ↓.

Mistake 2: Forgetting the “Weak vs Strong” Electrolyte Difference

The error:

Students treat all electrolytes the same — saying “κ always decreases linearly.”

Why it’s wrong:

For strong electrolytes (e.g., NaCl, HCl): κ decreases smoothly as dilution increases.
For weak electrolytes (e.g., CH₃COOH): κ decreases very sharply at first because dilution also increases dissociation (Le Chatelier’s principle), but the net effect is still a decrease — just not linear.

How to avoid:

Draw a rough graph in your mind:
- Strong: κ vs concentration → nearly straight line.
- Weak: κ drops steeply at low dilution, then flattens.
Remember: κ always decreases — the rate differs, not the direction.

Mistake 3: Saying “Conductivity Increases Because More Ions Are Produced”

The error:

Students think that adding water increases the total number of ions (true for weak electrolytes) and therefore κ increases.

Why it’s wrong:

Yes, for weak electrolytes, dilution shifts equilibrium to produce more ions per molecule.
But the volume increases even more.
Ion concentration (ions per cm³) still falls → κ falls.

How to avoid:

Use the concentration argument:

\kappa \propto \text{(number of ions per unit volume)}

Dilution → volume ↑ → concentration ↓ → κ ↓.

The increase in dissociation cannot compensate for the volume increase.

Mistake 4: Mixing Up “Conductance” and “Conductivity”

The error:

Students treat conductance (G) and specific conductivity (κ) as the same.

Why it’s wrong:

Conductance $G = \frac{1}{R}$ depends on cell geometry.
Specific conductivity $\kappa = G \times \frac{l}{A}$ is intensive — independent of cell size.
Adding water changes both, but the question asks about κ (intensive property).

How to avoid: …

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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 20 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$50 \Omega$
›Reveal solutionSolution
The key is to first find the cell constant from the 0.2 M data, then use it to find the conductivity of the 0.4 M solution, and finally compute its molar conductivity. The result matches option (C).

Concept and intuition

Molar conductivity ( $\Lambda_m$ ) is the conductivity of a solution divided by its molar concentration: $\Lambda_m = \frac{\kappa}{c}$ . But we cannot directly measure conductivity — we measure resistance $R$ of a solution in a cell with a fixed geometry. The cell constant $G^*$ (units m⁻¹) relates conductivity $\kappa$ to measured conductance $1/R$ :
$\kappa = G^* \cdot \frac{1}{R}$
So the plan: use the first solution’s known $\kappa$ and $R$ to find $G^*$ , then use that $G^*$ with the second solution’s $R$ to find its $\kappa$ , and finally compute $\Lambda_m$ for the second solution.

Step-by-step solution

Find the cell constant from the 0.2 M data For the 0.2 M solution:

$\kappa_1 = 1.3\ \mathrm{S\,m^{-1}}, \quad R_1 = 50\ \Omega$
Using $\kappa = G^* / R$ :
$G^* = \kappa_1 \cdot R_1 = 1.3 \times 50 = 65\ \mathrm{m^{-1}}$
This cell constant is fixed for the same conductivity cell.

Find the conductivity of the 0.4 M solution For the 0.4 M solution: $R_2 = 260\ \Omega$ .

$\kappa_2 = \frac{G^*}{R_2} = \frac{65}{260} = 0.25\ \mathrm{S\,m^{-1}}$

Compute the molar conductivity of the 0.4 M solution Molar conductivity $\Lambda_m$ is defined as:

$\Lambda_m = \frac{\kappa}{c}$
where $c$ is concentration in mol/m³. …
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COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{K}_{\mathrm{a}}$
›Reveal solutionSolution

The key idea is to combine Ostwald’s dilution law with the relation between molar conductivity and degree of dissociation. The correct expression is option (A).

We start from the fundamental relationship for a weak electrolyte: the degree of dissociation $\alpha$ is given by the ratio of molar conductivity at concentration $C$ to that at infinite dilution:
$\alpha = \frac{\lambda_m}{\lambda_m^\infty}$
This works because $\lambda_m$ increases with dissociation, and at infinite dilution the electrolyte is fully dissociated.

For a weak acid HA dissociating as:
$HA \rightleftharpoons H^+ + A^-$
Ostwald’s dilution law gives the ionization constant:
$K_a = \frac{C \alpha^2}{1 - \alpha}$
Now substitute $\alpha = \lambda_m / \lambda_m^\infty$ :
$K_a = \frac{C \left( \frac{\lambda_m}{\lambda_m^\infty} \right)^2}{1 - \frac{\lambda_m}{\lambda_m^\infty}} = \frac{C \lambda_m^2 / (\lambda_m^\infty)^2}{(\lambda_m^\infty - \lambda_m)/\lambda_m^\infty} = \frac{C \lambda_m^2}{\lambda_m^\infty (\lambda_m^\infty - \lambda_m)}$
This matches option (A) exactly. …
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COMEDK 2026Set 2026-M1 markMCQ
$Q. Two statements, one Assertion and the other Reason are given. Choose the right option. Assertion: The Molar conductivity of KCl increases very slowly with dilution and approaches a limiting value when dilution is infinite. Reason: In case of KCl there is an increase in the number of ions on dilution due to complete ionisation at infinite dilution. (A) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion (B) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion (C) Assertion is incorrect but Reason is correct (D) Assertion is correct but Reason is incorrect$
›Reveal solutionSolution
Molar conductivity increases with dilution due to reduced ion-ion interactions, not because more ions are formed — KCl is already fully dissociated. The Assertion is correct, but the Reason is wrong.

The key concept here is Kohlrausch’s law and the distinction between strong and weak electrolytes. KCl is a strong electrolyte — it dissociates completely in water at all dilutions. So the number of ions per formula unit does not increase as you dilute it; it’s already 100% ionised.

The Assertion says molar conductivity of KCl increases slowly with dilution and approaches a limiting value. That is true. But why does it increase? Not because more ions appear — they were already there. The increase happens because ion-ion interactions weaken as the solution gets more dilute. Ions move more freely, so conductivity per mole (molar conductivity) rises. At infinite dilution, ions are independent and molar conductivity reaches its maximum, the limiting molar conductivity $\Lambda_m^\infty$ .

The Reason claims that dilution causes an increase in the number of ions due to complete ionisation at infinite dilution. That is false for KCl — it is already completely ionised even at moderate concentrations. The number of ions per mole of KCl does not change with dilution. So the Reason is incorrect.

Let’s walk through it step by step.

Identify the electrolyte type. KCl is a salt of a strong acid (HCl) and a strong base (KOH). In water, it dissociates fully: $\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-$ . This is true at any dilution — there is no undissociated KCl present. So the degree of ionisation $\alpha = 1$ always.

Understand molar conductivity. Molar conductivity $\Lambda_m$ is defined as $\Lambda_m = \frac{\kappa}{c}$ , where $\kappa$ is conductivity and $c$ is concentration. For a strong electrolyte, $\Lambda_m$ increases as $c$ decreases, but the increase is gradual and follows the Kohlrausch square-root law: $\Lambda_m = \Lambda_m^\infty - A\sqrt{c}$ . The limiting value $\Lambda_m^\infty$ is approached as $c \to 0$ .

Why does $\Lambda_m$ increase? At higher concentrations, ions are close together and their mutual electrostatic attractions (ion-ion interactions) slow them down. Dilution separates the ions, reducing these interactions, so each ion moves faster under the applied field. The number of charge carriers per mole is constant — it’s always 2 moles of ions per mole of KCl — but their mobility increases. That’s the sole reason for the rise in $\Lambda_m$ . …
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KCET 2026Set D31 markMCQ
$\Lambda_m^o$
›Reveal solutionSolution

Apply Kohlrausch's law of independent ionic migration: combine strong electrolytes sharing NH $_4^+$ and OH $^-$ so the extra ions (Na $^+$ , Cl $^-$ ) cancel out.

Step 1 — Express each electrolyte's limiting conductivity in terms of ions
$\Lambda_m^o(\text{NH}_4\text{Cl}) = \lambda^o_{\text{NH}_4^+} + \lambda^o_{\text{Cl}^-}$ $\Lambda_m^o(\text{NaOH}) = \lambda^o_{\text{Na}^+} + \lambda^o_{\text{OH}^-}$ $\Lambda_m^o(\text{NaCl}) = \lambda^o_{\text{Na}^+} + \lambda^o_{\text{Cl}^-}$
Step 2 — Combine to isolate NH $_4^+$ and OH $^-$

Adding the first two and subtracting the third:

$\Lambda_m^o(\text{NH}_4\text{Cl}) + \Lambda_m^o(\text{NaOH}) - \Lambda_m^o(\text{NaCl}) = (\lambda^o_{\text{NH}_4^+} + \lambda^o_{\text{Cl}^-}) + (\lambda^o_{\text{Na}^+} + \lambda^o_{\text{OH}^-}) - (\lambda^o_{\text{Na}^+} + \lambda^o_{\text{Cl}^-})$ …
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COMEDK 2025Set 2025-A1 markMCQ
$\lambda_{\mathrm{m}}$
›Reveal solutionSolution
The assertion incorrectly states the slope is positive (+A) when it is actually negative for strong electrolytes; the reason correctly describes the trend but misattributes it to "slow increase" rather than a small, gradual decrease. The correct option is (C).

The key concept here is Kohlrausch’s law for strong electrolytes. For strong electrolytes (fully dissociated salts like NaCl, KCl), molar conductivity $\Lambda_m$ decreases linearly with the square root of concentration $\sqrt{c}$ , not increases. The slope is negative because as concentration increases, ion-ion interactions impede mobility. The reason, while describing the correct physical trend (conductivity changes with dilution), gets the direction wrong—it says "increases slowly" when in fact it decreases slightly with increasing concentration.

Let’s break it down step by step.

Understand the Assertion

The assertion says: For strong electrolytes, the plot of Molar conductivity versus Concentration gives a straight line with slope equal to +A and intercept equal to $\lambda_m$ .

Kohlrausch’s empirical law states: $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$ , where $\Lambda_m^0$ is the limiting molar conductivity (intercept) and $A$ is a positive constant.

This is a straight line when $\Lambda_m$ is plotted against $\sqrt{c}$ , not against $c$ directly. The slope is $-A$ (negative), not $+A$ .

Therefore, the assertion is incorrect because it gives the wrong sign for the slope and implies a plot against $c$ (not $\sqrt{c}$ ).

Understand the Reason

The reason says: For strong electrolytes, $\lambda_m$ increases slowly with dilution due to increase in the distance between the ions and increase in ionic mobility.

As we dilute a strong electrolyte, the distance between ions increases, reducing interionic attractions. This does increase ionic mobility, so $\Lambda_m$ increases.

However, the phrase "increases slowly" is misleading: for strong electrolytes, $\Lambda_m$ increases sharply at low concentrations and then approaches a constant $\Lambda_m^0$ ; the change is not "slow" in the sense of a small slope. More importantly, the reason correctly identifies the cause (greater distance → less hindrance → higher mobility), but it misstates the rate as "slow". …
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COMEDK 2025Set 2025-E1 markMCQ
$0.1 \mathrm{~mol} \mathrm{~L}^{-1}$
›Reveal solutionSolution
The key is to first find the cell constant from the known KCl data, then use it to get the conductivity of the unknown solution, and finally compute its molar conductivity. The result is $0.448 \, \mathrm{S \, m^2 \, mol^{-1}}$ , which matches option (C) when converted to $\mathrm{S \, cm^2 \, mol^{-1}}$ .

Concept & Intuition

Conductivity cells have a fixed geometry — the distance between electrodes and their area — captured by the cell constant $G^* = \frac{\ell}{A}$ (units: $\mathrm{m^{-1}}$ ).

We cannot measure $\ell$ and $A$ directly, but we can find $G^*$ using a standard solution of known conductivity.

Once we know $G^*$ , any unknown solution’s conductivity is simply $\kappa = G^* / R$ .

Then molar conductivity $\Lambda_m = \frac{\kappa}{c}$ (with careful unit conversion) gives the answer.

Step-by-step

Find the cell constant from the KCl data For the KCl solution:

$\kappa_{\text{KCl}} = 1.29 \, \mathrm{S \, m^{-1}}, \quad R_{\text{KCl}} = 80 \, \Omega$
The cell constant is:
$G^* = \kappa_{\text{KCl}} \times R_{\text{KCl}} = 1.29 \times 80 = 103.2 \, \mathrm{m^{-1}}$

Use the cell constant to find the conductivity of the unknown For the unknown solution:

$R_{\text{unknown}} = 92 \, \Omega$ $\kappa_{\text{unknown}} = \frac{G^*}{R_{\text{unknown}}} = \frac{103.2}{92} \approx 1.12174 \, \mathrm{S \, m^{-1}}$

Convert concentration to SI units The concentration is $0.025 \, \mathrm{M} = 0.025 \, \mathrm{mol \, L^{-1}}$ . Since $1 \, \mathrm{L} = 10^{-3} \, \mathrm{m^3}$ :
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COMEDK 2025Set 2025-M1 markMCQ
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$
›Reveal solutionSolution
Molar conductivity depends on the number of ions produced per formula unit in solution. The more ions, the higher the conductivity. The correct order is D > A > C > B, which corresponds to option (B).

The key concept here is Kohlrausch’s law of independent migration of ions: the molar conductivity of an electrolyte at infinite dilution is the sum of the conductivities of its constituent ions. For a given concentration (here, aqueous solutions at comparable dilution), the compound that dissociates into more ions will have a higher molar conductivity. So we simply count the number of ions each coordination compound releases when dissolved in water.

Identify the number of ions per formula unit for each compound.

Coordination compounds in water typically dissociate into the complex ion and the counter ions outside the coordination sphere. The ligands inside the brackets are tightly bound and do not dissociate.

A: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

The complex ion is $[\mathrm{Co(NH_3)_5Cl}]^{2+}$ and there are two $\mathrm{Cl}^-$ ions outside.

→ Total ions = 1 complex cation + 2 chloride ions = 3 ions.

B: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]$

All three chlorines are inside the coordination sphere; no counter ions outside.

→ This is a neutral complex, so it does not dissociate into ions.

→ Total ions = 0 ions (or effectively 1 molecule, but conductivity is negligible).

C: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$

The complex ion is $[\mathrm{Co(NH_3)_4Cl_2}]^{+}$ and one $\mathrm{Cl}^-$ outside.

→ Total ions = 1 complex cation + 1 chloride ion = 2 ions.

D: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$

The complex ion is $[\mathrm{Co(NH_3)_6}]^{3+}$ and three $\mathrm{Cl}^-$ outside. …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{CH}_3 \mathrm{COOH}$
›Reveal solutionSolution
The pKa is found by first calculating the molar conductivity and degree of dissociation from the given conductivity and limiting conductivities, then using the dissociation constant expression for a weak acid. The result is approximately 4.73, corresponding to option (A).

Concept & Intuition

This problem connects conductivity measurements to acid dissociation. For a weak acid like acetic acid, the molar conductivity at a given concentration is less than the limiting molar conductivity because the acid is only partially dissociated. The ratio of these gives the degree of dissociation ( $\alpha$ ). Once $\alpha$ is known, the dissociation constant $K_a$ follows from the equilibrium expression, and then $\mathrm{p}K_a = -\log K_a$ .

Calculate the molar conductivity ( $\Lambda_m$ ) of the solution. Molar conductivity is given by $\Lambda_m = \frac{\kappa}{c}$ , where $\kappa$ is the conductivity and $c$ is the concentration in mol/cm³ (since units must match). Here, $\kappa = 1.65 \times 10^{-4} \, \mathrm{S\,cm^{-1}}$ and $c = 0.01 \, \mathrm{mol\,L^{-1}} = 0.01 \times 10^{-3} \, \mathrm{mol\,cm^{-3}} = 10^{-5} \, \mathrm{mol\,cm^{-3}}$ .

$\Lambda_m = \frac{1.65 \times 10^{-4}}{10^{-5}} = 16.5 \, \mathrm{S\,cm^2\,mol^{-1}}$

Find the limiting molar conductivity ( $\Lambda_m^0$ ) of acetic acid. Using Kohlrausch’s law:

$\Lambda_m^0(\mathrm{CH_3COOH}) = \lambda^0(\mathrm{H^+}) + \lambda^0(\mathrm{CH_3COO^-})$
Given $\lambda^0(\mathrm{H^+}) = 349.1$ and $\lambda^0(\mathrm{CH_3COO^-}) = 40.9$ (both in $\mathrm{S\,cm^2\,mol^{-1}}$ ):
$\Lambda_m^0 = 349.1 + 40.9 = 390.0 \, \mathrm{S\,cm^2\,mol^{-1}}$

Determine the degree of dissociation ( $\alpha$ ). For a weak electrolyte, $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$ .

$\alpha = \frac{16.5}{390.0} \approx 0.04231$

Write the dissociation equilibrium and find $K_a$ . For $\mathrm{CH_3COOH} \rightleftharpoons \mathrm{H^+} + \mathrm{CH_3COO^-}$ , initial concentration $c = 0.01 \, \mathrm{M}$ . At equilibrium: $[\mathrm{H^+}] = [\mathrm{CH_3COO^-}] = c\alpha, \quad [\mathrm{CH_3COOH}] = c(1-\alpha) …$
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KCET 2024Set B-21 markMCQ
$\lambda_m = \lambda_m^0 - A\sqrt{C}$
›Reveal solutionSolution
The slope $A$ of the Debye–Hückel–Onsager plot is fixed by the electrolyte's charge type (1:1, 1:2, 2:2 …), so the pair with the same charge type shares the same $A$ .

Step 1 — What the equation says.

For a strong electrolyte, molar conductivity falls with concentration as
$\Lambda_m = \Lambda_m^{0} - A\sqrt{C}.$
The $\sqrt{C}$ dependence comes from the ionic atmosphere (relaxation + electrophoretic effects). The Debye–Hückel–Onsager theory shows the constant $A$ depends on:

the nature of the solvent (dielectric constant, viscosity),

the temperature, and

the charge type of the electrolyte (i.e. the valencies $z_+$ , $z_-$ ).

It does not depend on which particular ions of that charge type are present.

Step 2 — Classify each substance by charge type.

Electrolyte Ions Charge type
NaCl Na $^+$ , Cl $^-$ 1 : 1
KBr K $^+$ , Br $^-$ 1 : 1
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{NH}_4 \mathrm{OH}$
›Reveal solutionSolution
The degree of ionization is the ratio of molar conductivity at a given concentration to the limiting molar conductivity. For NH₄OH, this gives 9.54 / 238 ≈ 0.04008, which is 4.008%, so option (A) is correct.

The key idea here is Kohlrausch’s law and the concept of degree of ionization for weak electrolytes. Ammonium hydroxide (NH₄OH) is a weak base; it does not fully dissociate in water. The molar conductivity at a given concentration (Λₘ) is less than the limiting molar conductivity (Λₘ⁰) because only a fraction of the molecules are ionized. For weak electrolytes, the degree of ionization (α) is simply:
$\alpha = \frac{\Lambda_m}{\Lambda_m^0}$
This works because the mobility of ions is constant at infinite dilution, and the only reason Λₘ is smaller is that fewer ions are present.

Now, let’s work through it step by step.

Identify the given data

Limiting molar conductivity, $\Lambda_m^0 = 238 \, \text{S cm}^2 \text{mol}^{-1}$

Molar conductivity at 0.1 M, $\Lambda_m = 9.54 \, \text{S cm}^2 \text{mol}^{-1}$

Concentration = 0.1 M, temperature = 25°C

Recall the formula for degree of ionization

For a weak electrolyte, the degree of ionization is:

$\alpha = \frac{\Lambda_m}{\Lambda_m^0}$
This is valid because at infinite dilution, the electrolyte is fully dissociated, so Λₘ⁰ represents the conductivity if all molecules were ions. At any finite concentration, the actual conductivity is proportional to the fraction dissociated.

Plug in the numbers

$\alpha = \frac{9.54}{238}$
Calculate:
$\alpha = 0.040084...$

Convert to percentage

$\alpha \times 100\% = 0.040084 \times 100\% = 4.0084\%$
Rounded to three significant figures, this is 4.008%.

Match with the options …
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COMEDK 2024Set 2024-E1 markMCQ
$\sqrt{C}(\mathrm{~mol} / \mathrm{L})^{1 / 2}$
›Reveal solutionSolution
The assertion is correct (Kohlrausch’s law gives a straight line for strong electrolytes, with the same slope for a given type), but the reason is false (strong electrolytes dissociate completely at infinite dilution, so the number of ions is the same for a given type). Hence the correct option is (A).

Concept and Intuition

Kohlrausch discovered that for strong electrolytes, molar conductivity $\Lambda_m$ decreases linearly with the square root of concentration at low concentrations. This is described by $\Lambda_m = \Lambda_m^\infty - A\sqrt{C}$ , where $\Lambda_m^\infty$ is the limiting molar conductivity and $A$ is a constant that depends only on the type of electrolyte (e.g., 1:1, 2:1, etc.) and the solvent, not on the specific ions. The reason given claims that at infinite dilution, strong electrolytes of the same type have different numbers of ions due to incomplete dissociation — but that’s wrong: strong electrolytes dissociate completely at all dilutions, and at infinite dilution they are fully dissociated, so all 1:1 electrolytes give exactly two ions per formula unit, all 2:1 give three, etc. The reason therefore does not explain the assertion and is itself false.

Step-by-step reasoning

Understanding the assertion For a strong electrolyte, Kohlrausch’s law states:

$\Lambda_m = \Lambda_m^\infty - A\sqrt{C}$
This is a linear equation in $\sqrt{C}$ . The intercept (at $\sqrt{C}=0$ ) is $\Lambda_m^\infty$ , and the slope is $-A$ . The constant $A$ depends on the stoichiometry (type) of the electrolyte and the solvent properties (viscosity, dielectric constant), but is the same for all electrolytes of a given type (e.g., all 1:1 electrolytes like NaCl, KCl, HCl have the same $A$ in the same solvent). So the assertion is correct.

Examining the reason The reason says: “At infinite dilution, strong electrolytes of the same type will have different number of ions due to incomplete dissociation.”

Strong electrolytes (e.g., NaCl, KOH, HCl) dissociate completely in solution at all concentrations, including infinite dilution. …
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COMEDK 2024Set 2024-M1 markMCQ
$0.1 \mathrm{M}$
›Reveal solutionSolution
The molar conductivity is found by first calculating the conductivity from the cell geometry and measured current/voltage, then dividing by the concentration. The result is $95.5 \, \mathrm{S\,cm^2\,mol^{-1}}$ , which corresponds to option (B).

We are given a conductivity cell with a solution of $0.1\,\mathrm{M}$ $\mathrm{AgNO}_3$ . A voltage of $40\,\mathrm{V}$ is applied, producing a current of $0.4\,\mathrm{A}$ . The column has diameter $4.0\,\mathrm{cm}$ and length $12\,\mathrm{cm}$ . We need the molar conductivity $\Lambda_m$ .

Concept and intuition:

Molar conductivity is the conductivity of a solution divided by its molar concentration. Conductivity $\kappa$ itself is the reciprocal of resistivity, and we can get it from the measured resistance $R = V/I$ and the cell constant $G^* = \ell / A$ (length over cross-sectional area). So the plan is:

Find resistance from Ohm’s law.

Compute the cell constant from the given dimensions.

Conductivity $\kappa = \text{cell constant} / R$ .

Molar conductivity $\Lambda_m = \kappa / c$ (with careful unit conversion).

Let’s go step by step.

Find the resistance of the solution column. Ohm’s law: $R = V / I$ .

$R = \frac{40\,\mathrm{V}}{0.4\,\mathrm{A}} = 100\,\Omega.$

Compute the cell constant $G^*$ . The cell constant is $\ell / A$ , where $\ell$ is the distance between electrodes (length of column) and $A$ is the cross-sectional area. Diameter $d = 4.0\,\mathrm{cm}$ , so radius $r = 2.0\,\mathrm{cm}$ . Area:

$A = \pi r^2 = \pi (2.0)^2 = 4\pi \,\mathrm{cm}^2.$
Length $\ell = 12\,\mathrm{cm}$ .

Hence:
$G^* = \frac{\ell}{A} = \frac{12}{4\pi} = \frac{3}{\pi} \,\mathrm{cm}^{-1}.$

Calculate the conductivity $\kappa$ . Conductivity is related to resistance by:

$\kappa = \frac{G^*}{R}.$
So:
$\kappa = \frac{3/\pi}{100} = \frac{3}{100\pi} \,\mathrm{S\,cm^{-1}}.$
Numerically:
$\kappa \approx \frac{3}{314.159} \approx 0.009549\,\mathrm{S\,cm^{-1}}.$

Convert concentration to proper units. …
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Electrolyte	Ions	Charge type
NaCl	Na $^+$ , Cl $^-$	1 : 1
KBr	K $^+$ , Br $^-$	1 : 1

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Q.In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

Concept understanding — Molar Conductivity

From Resistance to Conductance: Flipping the Idea

Ohm's Law in Conductance Form

Now, Conductivity: The Material's Intrinsic Property

The Big Picture in One Table

Intuitive Analogy

Why this formula?

Conductance and Conductivity: Why the Formulas Hold

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

2. Ohm's Law in Terms of Conductance

3. From Resistance to Conductivity: The Geometry Factor

4. The Key Formula: Conductance in Terms of Conductivity

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

Why this happens — the core idea

Step-by-step reasoning

Method: Dilution Effect Analysis (Kohlrausch's Approach)

Step 1 – Understand the two opposing factors

Step 2 – Recall the definition

Step 3 – Apply the dilution logic

Step 4 – State the final trend …

Mistake 1: Confusing Specific Conductivity with Molar Conductivity

Mistake 2: Forgetting the “Weak vs Strong” Electrolyte Difference

Mistake 3: Saying “Conductivity Increases Because More Ions Are Produced”

Mistake 4: Mixing Up “Conductance” and “Conductivity”

Step 1 — Express each electrolyte's limiting conductivity in terms of ions

Step 2 — Combine to isolate NH $_4^+$ and OH $^-$

Concept & Intuition

Step-by-step

Q.In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

Why this happens — the core idea

Step-by-step reasoning

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