Lambda⁰m(NH_4OH) is equal to _______. (i) Lambda⁰m – Class 12 Chemistry

NCERT Exemplar · Q16

\Lambda^0_{m(NH_4OH)}

(i)

\Lambda^0_{m(NH_4OH)} + \Lambda^0_{m(NH_4Cl)} - \Lambda^0_{m(HCl)}

(ii)

\Lambda^0_{m(NH_4Cl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}

(iii)

\Lambda^0_{m(NH_4Cl)} + \Lambda^0_{m(NaCl)} - \Lambda^0_{m(NaOH)}

(iv)

\Lambda^0_{m(NaOH)} + \Lambda^0_{m(NaCl)} - \Lambda^0_{m(NH_4Cl)}

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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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The limiting molar conductivity of a weak electrolyte like $NH_4OH$ can be found by combining the $\Lambda_m^0$ values of strong electrolytes that share its ions. Using Kohlrausch’s law of independent ion migration, the correct expression is $\Lambda_{m(NH_4OH)}^0 = \Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$ , which corresponds to option (ii).

The key idea here is Kohlrausch’s law: at infinite dilution, each ion contributes a fixed amount to the molar conductivity, independent of the other ion it travels with. So the limiting molar conductivity of any electrolyte is simply the sum of the limiting conductivities of its constituent ions.

For a weak base like $NH_4OH$ , we cannot measure $\Lambda_m^0$ directly by extrapolation (because it doesn’t fully dissociate even at low concentrations). But we can build it from the $\Lambda_m^0$ values of strong electrolytes that contain the same ions — $NH_4^+$ and $OH^-$ — by adding and subtracting known values to cancel out the unwanted ions.

Let’s see how.

Write what we want in terms of ions.

$\Lambda_{m(NH_4OH)}^0 = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$

That’s our target.
Find strong electrolytes that give us these ions.
- $NH_4Cl$ gives $\lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0$
- $NaOH$ gives $\lambda_{Na^+}^0 + \lambda_{OH^-}^0$
- $NaCl$ gives $\lambda_{Na^+}^0 + \lambda_{Cl^-}^0$
Combine them to isolate the target sum.

If we add the first two and subtract the third:

(\lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0)

The $\lambda_{Na^+}^0$ and $\lambda_{Cl^-}^0$ cancel perfectly, leaving:

\lambda_{NH_4^+}^0 + \lambda_{OH^-}^0 = \Lambda_{m(NH_4OH)}^0

Translate back to electrolyte notation. So: …

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Common Mistakes & How to Avoid Them

Mistake 1: Not Understanding Kohlrausch’s Law

Students often try to memorise the answer without knowing why the formula works.

How to avoid:

Kohlrausch’s Law states that at infinite dilution, molar conductivity is the sum of independent ionic contributions:

\Lambda_m^0 = \lambda_+^0 + \lambda_-^0

For a weak base like $NH_4OH$ , you cannot measure $\Lambda_m^0$ directly (it doesn’t fully dissociate). So you build it from strong electrolytes whose $\Lambda_m^0$ values are known.

Mistake 2: Forgetting to Cancel Ions Correctly

Students pick options without checking if the unwanted ions cancel out.

How to avoid:

Write each electrolyte as its ions, then cancel common ions.

For correct option (ii):

\Lambda_{m(NH_4Cl)}^0 = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0

\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0

\Lambda_{m(NaCl)}^0 = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0

Now do:

(\lambda_{NH_4^+} + \lambda_{Cl^-}) + (\lambda_{Na^+} + \lambda_{OH^-}) - (\lambda_{Na^+} + \lambda_{Cl^-})

Cancel $\lambda_{Na^+}$ and $\lambda_{Cl^-}$ → you get:

\boxed{\lambda_{NH_4^+}^0 + \lambda_{OH^-}^0 = \Lambda_{m(NH_4OH)}^0}

Mistake 3: Confusing Addition/Subtraction Signs

Students misplace the minus sign, especially in options like (iii) or (iv).

How to avoid:

Always write the ionic breakdown before deciding the sign. The target is $NH_4^+ + OH^-$ .

You need $NH_4^+$ from $NH_4Cl$
You need $OH^-$ from $NaOH$
You must subtract the common ion pair ( $Na^+ + Cl^-$ ) that appears twice — that’s $NaCl$ .

So the correct structure is:

\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0

Mistake 4: Picking Option (i) Without Checking

Option (i) looks similar but uses $HCl$ instead of $NaOH$ and $NaCl$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04214 marksMCQ
$^2$
›Reveal solutionSolution

$\alpha=\Lambda_m/\Lambda_m^\circ=70/420=16.7\\%$ .

Limiting molar conductivity of HA:
$\Lambda_m^\circ = \lambda_{H^+}^\circ+\lambda_{A^-}^\circ = 340+80 = 420\\,Scm^2mol^{-1}.$
Degree of ionisation: …
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KEAM 2026Set pha-2026-0418F4 marksMCQ
$CaCl_{2}$
›Reveal solutionSolution

Kohlrausch: $\Lambda^\circ_m(CaCl_2) = \lambda^\circ(Ca^{2+}) + 2\lambda^\circ(Cl^-)$ .

$271.6 = 119 + 2\,\lambda^\circ(Cl^-)$ …
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KEAM 2025Set eng-2025-04254 marksMCQ
$\Lambda_m$
›Reveal solutionSolution

Degree of dissociation of a weak electrolyte $\alpha = \dfrac{\Lambda_m}{\Lambda_m^0} = \dfrac{78.1}{390.5} = 0.20$ .

For a weak electrolyte such as acetic acid, the degree of dissociation is the ratio of the molar conductivity at the given concentration to the molar conductivity at infinite dilution:

$\alpha = \frac{\Lambda_m}{\Lambda_m^0}$ …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$H^{+}$
›Reveal solutionSolution
The proton conducts by the Grotthuss (hopping) mechanism, giving $H^+$ by far the largest limiting molar conductivity ( $\approx 350\ \text{S cm}^2\ \text{mol}^{-1}$ ).

Limiting molar conductivities at 298 K (in $\text{S cm}^2\ \text{mol}^{-1}$ ) are roughly:

$H^+ \approx 349.8$

$K^+ \approx 73.5$

$Ca^{2+} \approx 119$ (but per mole of charge $\approx 59.5$ )

$Na^+ \approx 50.1$

$Mg^{2+} \approx 106$ (per mole of charge $\approx 53$ ) …
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KEAM 2024Set eng-2024-06074 marksMCQ
$^{-1}$
›Reveal solutionSolution

Molar conductivity $\Lambda_m=\kappa/c$ . With $\kappa=0.248$ S m $^{-1}$ and $c=0.02$ mol L $^{-1}=20$ mol m $^{-3}$ , $\Lambda_m=1.24\times10^{-2}$ S m $^2$ mol $^{-1}$ .

Molar conductivity is related to conductivity by
$\Lambda_m=\frac{\kappa}{c}.$
Using SI units, the concentration must be in mol m $^{-3}$ :

$c=0.02\ \text{mol L}^{-1}=0.02\times1000=20\ \text{mol m}^{-3}.$ …
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KEAM 2024Set eng-2024-06084 marksMCQ
$CH_3COONa$
›Reveal solutionSolution

Kohlrausch's law of independent ion migration lets the limiting molar conductance of acetic acid be built from those of sodium acetate, HCl and NaCl: $91.0+425.9-126.4=390.5$ S cm $^2$ mol $^{-1}$ .

Acetic acid dissociates into $CH_3COO^-$ and $H^+$ . Combining known limiting conductances:

$\Lambda_m^{\circ}(CH_3COOH)=\Lambda_m^{\circ}(CH_3COONa)+\Lambda_m^{\circ}(HCl)-\Lambda_m^{\circ}(NaCl).$ …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$^3$
›Reveal solutionSolution
The SI unit of molar conductivity is S m $^2$ mol $^{-1}$ .

Concept and Intuition

Molar conductivity is the conductivity per unit molar concentration, defined as $\Lambda_m = \kappa / c$ . Its unit follows directly from the units of conductivity and concentration in SI.

Step-by-Step Solution

Conductivity $\kappa$ has SI unit S m $^{-1}$ .

Concentration $c$ in SI is mol m $^{-3}$ . …
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Q. $\Lambda^0_{m(NH_4OH)}$ is equal to ______________.

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)

Method: Kohlrausch’s Law of Independent Migration of Ions

Steps

Common Mistakes & How to Avoid Them

Mistake 1: Not Understanding Kohlrausch’s Law

Mistake 2: Forgetting to Cancel Ions Correctly

Mistake 3: Confusing Addition/Subtraction Signs

Mistake 4: Picking Option (i) Without Checking

Concept and Intuition

Step-by-Step Solution

Q.Λm(NH4OH)0\Lambda^0_{m(NH_4OH)}Λm(NH4​OH)0​ is equal to ______________.

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Q. $\Lambda^0_{m(NH_4OH)}$ is equal to ______________.

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