Lambda⁰m(H_2O) is equal to ________. (Two or more than two options may be correct.) (i) Lambda⁰m(HCl) + Lambda

NCERT Exemplar · Q47

\Lambda^0_{m(H_2O)}

(i)

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

(ii)

\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaNO_3)} - \Lambda^0_{m(NaOH)}

(iii)

\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaNO_3)}

(iv)

\Lambda^0_{m(NH_4OH)} + \Lambda^0_{m(HCl)} - \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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Concept: Molar Conductivity at infinite dilution — Kohlrausch’s law states that $\Lambda_m^0$ of an electrolyte is the sum of the limiting molar conductivities of its constituent ions. For water, $\Lambda_{m(H_2O)}^0 = \lambda_{H^+}^0 + \lambda_{OH^-}^0$ . Only combinations built from strong electrolytes count, since a strong electrolyte's $\Lambda_m^0$ is the only kind that can be measured directly.

We need to combine strong electrolytes so that the net ionic sum equals $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ .

Step 1: Write the ionic contributions for each option.

For (i):

$\Lambda_{m(HCl)}^0 = \lambda_{H^+}^0 + \lambda_{Cl^-}^0$

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

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

Sum: $(\lambda_{H^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) = \lambda_{H^+}^0 + \lambda_{OH^-}^0$ ✓

Step 2: Check (ii):

$\Lambda_{m(HNO_3)}^0 = \lambda_{H^+}^0 + \lambda_{NO_3^-}^0$

$\Lambda_{m(NaNO_3)}^0 = \lambda_{Na^+}^0 + \lambda_{NO_3^-}^0$

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

Sum: $(\lambda_{H^+}^0 + \lambda_{NO_3^-}^0) + (\lambda_{Na^+}^0 + \lambda_{NO_3^-}^0) - (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) = \lambda_{H^+}^0 + 2\lambda_{NO_3^-}^0 - \lambda_{OH^-}^0$ ✗

Step 3: Check (iii):

$\Lambda_{m(HNO_3)}^0 = \lambda_{H^+}^0 + \lambda_{NO_3^-}^0$

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

$\Lambda_{m(NaNO_3)}^0 = \lambda_{Na^+}^0 + \lambda_{NO_3^-}^0$ …

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The limiting molar conductivity of water, $\Lambda^0_{m(H_2O)}$ , is found by applying Kohlrausch’s law of independent migration of ions. It equals the sum of the limiting conductivities of its constituent ions, $H^+$ and $OH^-$ , which can be obtained by combining the conductivities of strong electrolytes. The correct expressions are (i) and (iii).

The key idea here is Kohlrausch’s law: at infinite dilution, each ion contributes a fixed amount to the molar conductivity of an electrolyte, independent of the other ion it travels with. So $\Lambda^0_m$ for any electrolyte is simply the sum of the limiting conductivities of its cation and anion.

For water, which dissociates as $H_2O \rightleftharpoons H^+ + OH^-$ , its limiting molar conductivity is:

\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}

We don’t know these individual ionic conductivities directly, but we can get them by combining data from strong electrolytes that contain these ions — strong electrolytes are the ones whose $\Lambda_m^0$ can actually be measured directly, by extrapolating $\Lambda_m$ vs $\sqrt{c}$ to zero concentration.

Let’s check each option step by step.

Option (i): $\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}$

Write each in terms of ionic conductivities:
- $\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}$
Adding the first two and subtracting the third:

(\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-})

The $\lambda^0_{Na^+}$ and $\lambda^0_{Cl^-}$ cancel, leaving $\lambda^0_{H^+} + \lambda^0_{OH^-}$ , which is exactly $\Lambda^0_{m(H_2O)}$ . HCl, NaOH and NaCl are all strong electrolytes, so this is a legitimate calculation. (i) is correct.

Option (ii): $\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaNO_3)} - \Lambda^0_{m(NaOH)}$

Write them out:
- $\Lambda^0_{m(HNO_3)} = \lambda^0_{H^+} + \lambda^0_{NO_3^-}$
- $\Lambda^0_{m(NaNO_3)} = \lambda^0_{Na^+} + \lambda^0_{NO_3^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
Sum the first two and subtract the third:

(\lambda^0_{H^+} + \lambda^0_{NO_3^-}) + (\lambda^0_{Na^+} + \lambda^0_{NO_3^-}) - (\lambda^0_{Na^+} + \lambda^0_{OH^-})

The $\lambda^0_{Na^+}$ cancels, but we get $\lambda^0_{H^+} + 2\lambda^0_{NO_3^-} - \lambda^0_{OH^-}$ , which is not $\lambda^0_{H^+} + \lambda^0_{OH^-}$ . So (ii) is incorrect.

Option (iii): $\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaNO_3)}$

Write:
- $\Lambda^0_{m(HNO_3)} = \lambda^0_{H^+} + \lambda^0_{NO_3^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(NaNO_3)} = \lambda^0_{Na^+} + \lambda^0_{NO_3^-}$
Adding the first two and subtracting the third:

(\lambda^0_{H^+} + \lambda^0_{NO_3^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{NO_3^-})

The $\lambda^0_{Na^+}$ and $\lambda^0_{NO_3^-}$ cancel, leaving $\lambda^0_{H^+} + \lambda^0_{OH^-}$ . HNO₃, NaOH and NaNO₃ are all strong electrolytes, so this is also legitimate. (iii) is correct.

Option (iv): $\Lambda^0_{m(NH_4OH)} + \Lambda^0_{m(HCl)} - \Lambda^0_{m(NH_4Cl)}$ $Λ_{m (N H_{4} O H)}^{0} + Λ_{m (H C l)}^{0} - Λ_{m (N H_{4} C l)}^{0}$ Write:
- $\Lambda^0_{m(NH_4OH)} = \lambda^0_{NH_4^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$
- $\Lambda^0_{m(NH_4Cl)} = \lambda^0_{NH_4^+} + \lambda^0_{Cl^-}$ …

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Method: Kohlrausch’s Law of Independent Migration of Ions

Concept: At infinite dilution, each ion contributes a fixed amount to the molar conductivity of an electrolyte, independent of the other ion it is paired with. This lets you build $\Lambda_m^0$ of one substance from other substances' $\Lambda_m^0$ values — provided every substance used is a strong electrolyte (only strong-electrolyte $\Lambda_m^0$ can be measured directly, by extrapolating $\Lambda_m$ vs $\sqrt{c}$ to $c=0$ ).

Steps:

Write the expression for $\Lambda^0_m$ of water Water dissociates as:

H_2O \rightleftharpoons H^+ + OH^-

So,

\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}

Express each given electrolyte in terms of ionic conductivities

For example:
- $\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}$
Combine to isolate $\lambda^0_{H^+} + \lambda^0_{OH^-}$

Take option (i):

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

Substitute:

= (\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-})

Cancel $\lambda^0_{Na^+}$ and $\lambda^0_{Cl^-}$ :

= \lambda^0_{H^+} + \lambda^0_{OH^-} = \Lambda^0_{m(H_2O)}

HCl, NaOH, NaCl are all strong electrolytes — valid.

Check other options similarly
- Option (iii) also works, using only strong electrolytes HNO₃, NaOH, NaNO₃: …

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Common Mistakes & How to Avoid Them

Mistake 1: Not recognising Kohlrausch’s Law of independent migration of ions

Students often try to memorise the formula without understanding the logic behind it.

How to avoid:

Kohlrausch’s law says:

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

For water ( $H_2O$ ), the ions are $H^+$ and $OH^-$ . So:

\Lambda_{m(H_2O)}^0 = \lambda_{H^+}^0 + \lambda_{OH^-}^0

Now, any combination of strong electrolytes that gives you $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ is correct.

Mistake 2: Confusing addition/subtraction signs

Students often misplace the signs when combining electrolytes.

How to avoid:

Write each electrolyte’s ionic contributions explicitly:

$\Lambda_{m(HCl)}^0 = \lambda_{H^+}^0 + \lambda_{Cl^-}^0$
$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$
$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0$

Now compute:

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

Cancel $\lambda_{Na^+}^0$ and $\lambda_{Cl^-}^0$ → $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ ✓

Mistake 3: Assuming only one combination is correct

The question says two or more options may be correct, but students often stop after finding one.

How to avoid:

Check every option using the same method. For option (iii):

\Lambda_{m(HNO_3)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaNO_3)}^0

$\Lambda_{m(HNO_3)}^0 = \lambda_{H^+}^0 + \lambda_{NO_3^-}^0$
$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$
$\Lambda_{m(NaNO_3)}^0 = \lambda_{Na^+}^0 + \lambda_{NO_3^-}^0$

Cancel $\lambda_{Na^+}^0$ and $\lambda_{NO_3^-}^0$ → $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ ✓

So both (i) and (iii) are correct.

Mistake 4: Wrongly accepting option (iv) because the ion algebra 'cancels'

Option (iv) is $\Lambda_{m(NH_4OH)}^0 + \Lambda_{m(HCl)}^0 - \Lambda_{m(NH_4Cl)}^0$ . If you write it out formally:

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

This LOOKS identical to (i) and (iii) — but it is not accepted as a valid answer, and it's a genuine trap.

Why it's wrong: …

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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 32 on this concept.

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 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 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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CBSE 2025Set ANNUAL1 markMCQ
$Q. The unit of molar conductivity is$
(a) S cm^-2 mol^-1
(b) S cm^2 mol^-1
(c) S^-1 cm^2 mol^-1
(d) S cm^2 mol
›Reveal solutionSolution

Molar conductivity relates conductivity (S/cm) to concentration (mol/cm^3), giving the composite unit S cm^2 mol^-1.

Molar conductivity is defined as:

$\Lambda_m = \dfrac{\kappa \times 1000}{C}$

where $\kappa$ (specific conductivity) has units S cm^-1 and $C$ (concentration) has units mol L^-1 (mol per 1000 cm^3).

…

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CBSE 2025Set ANNUAL1 markMCQ
$L^{-1}$
(a) 12.9 S cm $^2$ mol $^{-1}$
(b) 1.29 S cm $^2$ mol $^{-1}$
(c) 0.0129 S cm $^2$ mol $^{-1}$
(d) 129 S cm $^2$ mol $^{-1}$
›Reveal solutionSolution

Converting conductivity (per cm) into molar conductivity requires dividing by the molar concentration expressed per cm³, giving a factor of 1000 in the standard formula.

Molar conductivity is related to the specific conductivity (electrolytic conductivity, $\kappa$ ) and molar concentration $C$ (in $mol\,L^{-1}$ ) by:

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

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The formula used to calculate molar conductivity of an electrolyte is _____.$
(a) $\Lambda=\dfrac{1000c}{k}$
(b) $c=\dfrac{1000\Lambda}{k}$
(c) $\Lambda=\dfrac{1000k}{c}$
(d) $k=\dfrac{1000}{\Lambda c}$
›Reveal solutionSolution

$\Lambda_m = \dfrac{1000\,\kappa}{c}$ .

Molar conductivity ( $\Lambda_m$ ) relates to specific conductivity ( $\kappa$ ) and molar concentration $c$ (in $mol\,dm^{-3}$ ) by:

$\Lambda_m = \dfrac{\kappa \times 1000}{c}$

…

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CBSE 2025Set ANNUAL1 mark
$Q. What is the SI unit of molar conductivity? OR Write the relation between specific conductivity and molar conductivity.$
›Reveal solutionSolution

Molar conductivity's SI unit follows from $\Lambda_m = \kappa/C$ : siemens metre-squared per mole.

Molar conductivity is defined as $\Lambda_m = \dfrac{\kappa}{C}$ , where $\kappa$ (specific/electrical conductivity) has SI unit $S\,m^{-1}$ and $C$ (molar concentration) has SI unit $mol\,m^{-3}$ . Dividing, the SI unit of $\Lambda_m$ works out to

$\frac{S\,m^{-1}}{mol\,m^{-3}} = S\,m^2\,mol^{-1}.$

(In practical lab work, where $\kappa$ is often expressed in $S\,cm^{-1}$ and $C$ in $mol\,L^{-1}$ , the commonly used relation is $\Lambda_m = \dfrac{1000\,\kappa}{C}$ , giving the c.g.s.-style unit $S\,cm^2\,mol^{-1}$ .)

…

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CBSE 2024Set A11 markMCQ
$Q. When the concentration of electrolytic solution approaches zero, the resulting molar conductivity is known as ;$
(a) specific conductance
(b) resistivity
(c) conductivity
(d) limiting molar conductivity
›Reveal solutionSolution

Molar conductivity at zero concentration (infinite dilution) is called the limiting molar conductivity — option (d).

Molar conductivity $\Lambda_m$ increases as an electrolytic solution is diluted, because more of the electrolyte is present as free, effectively conducting ions. As the concentration approaches zero (infinite dilution), $\Lambda_m$ reaches a limiting maximum value denoted $\Lambda_m^\circ$ , the **limiting molar co …
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CBSE 2024Set B1 mark
$Q. Answer in one word/sentence: Write the unit of Equivalence conductivity.$
›Reveal solutionSolution

Equivalent conductivity is conductivity per gram-equivalent of electrolyte per unit volume, giving units of S cm^2 eq^-1.

Equivalent conductivity, $\Lambda_{eq}$ , is defined as $\Lambda_{eq} = \kappa \times V$ , where $\kappa$ (specific conductivity) has units of S cm^-1 (ohm^-1 cm^-1) and V is the volume in cm^3 containing one gram-equivalent of …

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CBSE 2024Set ANNUAL1 mark
$2.5 \times 10^{-2}$
›Reveal solutionSolution

Adding the limiting ionic molar conductivities gives Λm°, and dividing the measured Λm by it gives the degree of dissociation.

For a weak electrolyte, the degree of dissociation is given by:

$\alpha = \frac{\Lambda_m}{\Lambda_m°}$

Limiting molar conductivity of methanoic acid (using Kohlrausch's law of independent migration of ions):

$\Lambda_m° = \lambda°_{H^+} + \lambda°_{HCOO^-} = 349.6 + 54.6 = 404.2\ S\,cm^2\,mol^{-1}$ …

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CBSE 2024Set ANNUAL1 mark
$Q. Write SI unit of molar conductivity.$
›Reveal solutionSolution

SI unit of molar conductivity is $S\,m^2\,mol^{-1}$ .

Molar conductivity $\Lambda_m$ is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte, measured between electrodes 1 m apart.

…

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Q.Λm(H2O)0\Lambda^0_{m(H_2O)}Λm(H2​O)0​ is equal to _______________. (Two or more than two options may be correct.)

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Q. $\Lambda^0_{m(H_2O)}$ is equal to _______________. (Two or more than two options may be correct.)

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