Match the terms given in Column I with the units given in Column II. Column I: (i) Lambdam (ii) ECell (iii) ka

NCERT Exemplar · Q53

Q. Match the terms given in Column I with the units given in Column II. Column I:

(i)

\Lambda_m

(ii)

E_{Cell}

(iii)

\kappa

(iv)

G^*

Column II:

(a)

S\ cm^{-1}

(b)

m^{-1}

(c)

S\ cm^2\ mol^{-1}

(d)

V

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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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This is a matching problem linking four electrochemical quantities to their correct SI or practical units. The key is to recall the definitions: molar conductivity $\Lambda_m$ is conductivity per mole, so its unit is $S\ cm^2\ mol^{-1}$ ; cell potential $E_{Cell}$ is voltage, so its unit is $V$ ; conductivity $\kappa$ is the reciprocal of resistivity, so its unit is $S\ cm^{-1}$ ; and the cell constant $G^*$ is the ratio of length to area, so its unit is $m^{-1}$ . The correct matches are: (i)→(c), (ii)→(d), (iii)→(a), (iv)→(b).

The question asks you to match four fundamental electrochemical quantities with their units. This is not about memorising a table — it is about understanding what each quantity physically represents. Once you know the definition, the unit follows naturally.

Let us go through each one.

1. Molar conductivity $\Lambda_m$

Molar conductivity is defined as the conductivity of a solution divided by the molar concentration of the electrolyte:

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

Here $\kappa$ has units of $S\ cm^{-1}$ (siemens per centimetre) and $c$ has units of $mol\ cm^{-3}$ (or more commonly $mol\ L^{-1}$ , but we must be careful with volume units). In practice, $\kappa$ is measured in $S\ cm^{-1}$ and concentration in $mol\ cm^{-3}$ , so:

\text{Unit of } \Lambda_m = \frac{S\ cm^{-1}}{mol\ cm^{-3}} = S\ cm^2\ mol^{-1}

That matches option (c).

Tip

A common shortcut: molar conductivity always has "per mole" in the denominator and "area" in the numerator — think of it as conductivity per unit concentration, which gives $S\ cm^2\ mol^{-1}$ .

2. Cell potential $E_{Cell}$

The electromotive force (emf) of a cell is simply the potential difference between its two electrodes. The SI unit of electric potential is the volt ( $V$ ). So $E_{Cell}$ is measured in volts. That matches option (d).

3. Conductivity $\kappa$

Conductivity is the reciprocal of resistivity ( $\rho$ ). Resistivity has units of $\Omega\ cm$ (ohm-centimetre), so:

\kappa = \frac{1}{\rho} \quad \Rightarrow \quad \text{Unit of } \kappa = \frac{1}{\Omega\ cm} = S\ cm^{-1}

Here $S$ (siemens) is the unit of conductance, equal to $\Omega^{-1}$ . So $\kappa$ has units $S\ cm^{-1}$ , matching option (a). …

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Method: Unit Analysis by Dimensional Matching

This method works by recalling the defining formula for each term, then deriving its unit from the SI/base units of the quantities in the formula.

Step 1: Recall the definition of each term

Term	Definition / Formula	Key quantities involved
$\Lambda_m$ (Molar conductivity)	$\Lambda_m = \frac{\kappa}{c}$	$\kappa$ = conductivity, $c$ = concentration (mol/cm³)
$E_{Cell}$ (Cell potential)	$E = -\frac{\Delta G}{nF}$	Voltage = J/C = V
$\kappa$ (Conductivity)	$\kappa = \frac{1}{\rho}$ (reciprocal of resistivity)	Unit: S/cm or S/m
$G^*$ (Cell constant)	$G^* = \frac{l}{A}$	$l$ = length, $A$ = area → unit: cm⁻¹ or m⁻¹

Step 2: Derive the unit for each

$\Lambda_m$ :

$\kappa$ has unit $S\,cm^{-1}$ , $c$ has unit $mol\,cm^{-3}$

$\Rightarrow \Lambda_m = \frac{S\,cm^{-1}}{mol\,cm^{-3}} = S\,cm^2\,mol^{-1}$

→ matches (c)
$E_{Cell}$ :

Directly measured in volts

→ matches (d)
$\kappa$ :

Conductivity unit is siemens per centimetre …

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Here are the common mistakes students make when matching these electrochemical quantities with their units, and how to avoid each.

Mistake 1: Confusing $\Lambda_m$ (Molar Conductivity) with $\kappa$ (Conductivity)

The error:

Students often swap the units of molar conductivity ( $\Lambda_m$ ) and specific conductivity ( $\kappa$ ). They match $\Lambda_m$ with $S\ cm^{-1}$ (conductivity) and $\kappa$ with $S\ cm^2\ mol^{-1}$ (molar conductivity).

Why it happens:

Both terms sound similar and both involve "conductivity." The formula $\Lambda_m = \frac{\kappa}{C}$ makes it easy to mix them up.

How to avoid:

Remember the definition:

Molar conductivity ( $\Lambda_m$ ) is the conductivity of a solution containing 1 mole of electrolyte. Its unit must involve per mole → $S\ cm^2\ mol^{-1}$ .
Conductivity ( $\kappa$ ) is the conductance of a 1 cm cube of solution → unit is $S\ cm^{-1}$ .

Trick: The "molar" in $\Lambda_m$ means "per mole" → look for $mol^{-1}$ in the unit.

Mistake 2: Forgetting that $E_{Cell}$ is measured in Volts

The error:

Some students match $E_{Cell}$ with $S\ cm^{-1}$ or $m^{-1}$ , thinking it is a conductivity-related term.

Why it happens:

$E_{Cell}$ appears alongside conductivity terms, so students assume it must also be a conductivity unit.

How to avoid:

Recall that $E$ stands for electromotive force (emf) or cell potential. The unit of potential difference is always the volt (V). No other unit in Column II has $V$ .

Key fact: $E_{Cell} = E^\circ_{cathode} - E^\circ_{anode}$ → result is in volts.

Mistake 3: Confusing $G^*$ (Cell Constant) with Conductivity

The error:

Students match $G^*$ with $S\ cm^{-1}$ or $S\ cm^2\ mol^{-1}$ , thinking it is a type of conductance.

Why it happens:

The symbol $G$ is often used for conductance, so $G^*$ looks like it should have a conductance unit.

How to avoid:

Know the definition:

$G^*$ (cell constant) = $\frac{l}{A}$ , where $l$ is length and $A$ is area. Its unit is $cm^{-1}$ or $m^{-1}$ . It is not a conductance — it is a geometric factor.

Trick: The star (*) means it is a constant for the cell, not a measured quantity. Units are inverse length.

Mistake 4: Mixing up $cm^{-1}$ and $m^{-1}$

The error:

Students match $G^*$ with $S\ cm^{-1}$ (conductivity) instead of $m^{-1}$ . …

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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 19 on this concept.

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$5 \times 10^{-5}\ S\ cm^{-1}$
›Reveal solutionSolution
Molar conductivity at 0.001 M works out to 50 S cm² mol⁻¹; dividing by the limiting value 390 gives the degree of dissociation, 0.128.

Concept and Intuition

For a weak electrolyte, the degree of dissociation can be estimated (Arrhenius) as the ratio of the molar conductivity at the given concentration to the molar conductivity at infinite dilution: $\alpha = \Lambda_m/\Lambda_m^\circ$ . This works because $\Lambda_m^\circ$ represents complete dissociation, so the ratio directly measures what fraction has actually dissociated.

Step-by-Step Solution

Convert $\kappa$ to $\Lambda_m$ : $\Lambda_m = \dfrac{\kappa\times1000}{C(\text{mol/L})}$ .

$\Lambda_m = \dfrac{5\times10^{-5}\times1000}{0.001} = \dfrac{5\times10^{-2}}{10^{-3}} = 50\ \mathrm{S\ cm^2\ mol^{-1}}$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$\sqrt{c}$
›Reveal solutionSolution
KCl and CsCl are the two close top curves (A and C), NaCl is the lower separate curve (B): option (B).

Concept and Intuition

For a strong electrolyte, $\Lambda_m = \Lambda^0_m - b\sqrt{c}$ , so the intercept at $\sqrt{c}\to0$ is the limiting molar conductivity $\Lambda^0_m = \lambda^0_{+} + \lambda^0_{-}$ . Because $Cl^-$ is shared, the curves are ordered purely by cation conductivity.

Step-by-Step Solution

Take $\lambda^0(Cl^-)\approx 76\ S\,cm^2\,mol^{-1}$ .

$\Lambda^0_m(NaCl)=50+76=126$ ; $\Lambda^0_m(KCl)=73+76=149$ ; $\Lambda^0_m(CsCl)=77+76=153$ .

KCl (149) and CsCl (153) differ by only $\sim4$ , so their lines nearly coincide — these are the overlapping pair ending at the adjacent points A and C.

NaCl (126) is far lower/apart from the pair — it is the distinct line ending at B.

Within the close pair, the slightly higher curve (C, above) is the larger $\Lambda^0_m$ = CsCl; the slightly lower (A) = KCl.

Common Mistakes …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$\Omega$
›Reveal solutionSolution
Use the KCl data to find the cell constant, then use that fixed cell constant to get the conductivity — and hence molar conductivity — of the $K_2SO_4$ solution. The answer is $3.4\times10^{-2}$ S m² mol⁻¹.

Concept and Intuition

A conductivity cell's geometry fixes a constant, the cell constant $G^* = \kappa \times R$ , which is the same for any solution measured in that same cell. Once $G^*$ is known from one calibrating solution (here KCl), it can be used to convert any other measured resistance directly into that solution's conductivity, from which molar conductivity follows by dividing by molar concentration.

Step-by-Step Solution

From the KCl data: $G^* = \kappa \times R = 0.3\ \text{S/m} \times 85\ \Omega = 25.5\ \text{m}^{-1}$ .

This same $G^*$ applies to the $K_2SO_4$ measurement in the same cell: $\kappa_{K_2SO_4} = \dfrac{G^*}{R} = \dfrac{25.5}{300} = 0.085\ \text{S/m}$ .

Convert concentration to SI (mol/m³): $0.0025\ \text{mol/L} = 2.5\ \text{mol/m}^3$ . …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$\Lambda_m$
›Reveal solutionSolution
Strong electrolytes like $NaCl$ and $CsCl$ show $\Lambda_m$ falling linearly with $\sqrt{c}$ (Kohlrausch's law); since $Cs^+$ has a higher limiting ionic conductivity than $Na^+$ , the $CsCl$ line sits above the $NaCl$ line, and the two lines run roughly parallel rather than crossing.

Concept and Intuition

Kohlrausch found that for a strong electrolyte, the molar conductivity varies with concentration as $\Lambda_m = \Lambda^0_m - A\sqrt{c}$ — a straight line with a negative slope when plotted against $\sqrt{c}$ , extrapolating to the limiting molar conductivity $\Lambda^0_m$ at $c\to0$ (i.e. at $\sqrt{c}=0$ ). This is very different from weak electrolytes, whose $\Lambda_m$ rises steeply (near-vertically) close to $c=0$ and cannot be extrapolated linearly.

The limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its ions (Kohlrausch's law of independent migration): $\Lambda^0_m(NaCl) = \lambda^0(Na^+) + \lambda^0(Cl^-)$ and $\Lambda^0_m(CsCl) = \lambda^0(Cs^+) + \lambda^0(Cl^-)$ . Since both share the same $Cl^-$ contribution, and $\lambda^0(Cs^+) = 77 > \lambda^0(Na^+) = 50$ , we must have $\Lambda^0_m(CsCl) > \Lambda^0_m(NaCl)$ — the $CsCl$ line intercepts the y-axis higher than the $NaCl$ line.

Step-by-Step Solution

Both $NaCl$ and $CsCl$ are strong electrolytes → both lines must be straight with negative slope (rules out options with positive slope or horizontal lines).

$\Lambda^0_m(CsCl) > \Lambda^0_m(NaCl)$ since $\lambda^0(Cs^+) > \lambda^0(Na^+)$ → the $CsCl$ line must lie above the $NaCl$ line, especially at (and near) $\sqrt{c}=0$ . …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$0.0115\ \mathrm{S\, cm^{-1}}$
›Reveal solutionSolution
Converting specific conductance to molar conductance using $\Lambda_m = 1000\kappa/C$ gives 230 S cm² mol⁻¹. Answer: (D).

Concept and Intuition

Specific conductance ( $\kappa$ ) is the conductance of a 1 cm cube of solution, while molar conductance ( $\Lambda_m$ ) normalizes this to "per mole of electrolyte," accounting for the fact that the amount of electrolyte (and hence total ions) present per unit volume depends on the molar concentration. The conversion factor of 1000 arises because $\kappa$ is defined per cm³ but concentration is given per litre (1000 cm³).

Step-by-Step Solution

Formula: $\Lambda_m = \dfrac{\kappa \times 1000}{C}$ , where $\kappa$ is in S cm⁻¹ and $C$ is molar concentration in mol L⁻¹.

Given $\kappa = 0.0115$ S cm⁻¹ and $C = 0.05$ mol L⁻¹.

$\Lambda_m = \dfrac{0.0115 \times 1000}{0.05} = \dfrac{11.5}{0.05}$ . …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$\mathrm{KCl}$
›Reveal solutionSolution
First get the (fixed) cell constant from the 0.1 M KCl calibration measurement, then use it with the second solution's resistance to get its conductivity, then convert to molar conductivity. The answer is (A) 124 S cm² mol⁻¹.

Concept and Intuition

A conductivity cell has a fixed geometric cell constant $G^* = \ell/A$ (length between electrodes / cross-sectional area), which relates measured resistance to the solution's conductivity: $\kappa = G^*/R$ , i.e. $G^* = \kappa R$ . Since $G^*$ is a property of the cell, not the solution, once we calibrate it with one known solution (0.1 M KCl of known conductivity), we can use the same $G^*$ for any other solution measured in that same cell — here, the 0.02 M KCl solution. Molar conductivity then normalises conductivity by concentration: $\Lambda_m = \dfrac{\kappa \times 1000}{C}$ (with $\kappa$ in S/cm and $C$ in mol/L, giving $\Lambda_m$ in S cm² mol⁻¹).

Step-by-Step Solution

Calibrate the cell constant using the 0.1 M KCl data: $\kappa_1 = 1.29\ \mathrm{S\,m^{-1}}$ , $R_1 = 100\ \Omega$ .

$G^* = \kappa_1 R_1 = 1.29 \times 100 = 129\ \mathrm{m^{-1}}$

Use the same cell (same $G^*$ ) with the second solution: $R_2 = 520\ \Omega$ . $\kappa_2 = \frac{G^*}{R_2} = \frac{129}{520} = 0.2481\ \mathrm{S\,m^{-1}}$ …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\Lambda_m$
›Reveal solutionSolution
Matching electrical/conductance quantities to their SI/CGS units: $\Lambda_m\to S\,cm^2\,mol^{-1}$ , $G\to S$ , $\kappa\to S\,cm^{-1}$ , $G^{*}\to m^{-1}$ — option (B).

Concept and Intuition

Electrolytic conductance quantities form a small family that is easy to confuse: conductance $G$ is simply the reciprocal of resistance and carries the base unit siemens; specific conductance (conductivity) $\kappa$ is conductance normalised per unit cell geometry, giving $S\,cm^{-1}$ ; molar conductivity $\Lambda_m$ further normalises by concentration, giving $S\,cm^2\,mol^{-1}$ ; and the cell constant $G^{*}=l/A$ is a pure geometric factor with units of reciprocal length.

Step-by-Step Solution

$\Lambda_m$ (molar conductivity) $= \dfrac{\kappa \times 1000}{C}$ , units work out to $S\,cm^2\,mol^{-1}$ → matches III.

$G$ (conductance) $=1/R$ , base SI unit siemens (S) → matches IV.

$\kappa$ (specific conductance/conductivity) has units $S\,cm^{-1}$ → matches I. …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$\mathrm{dm^{-3}}$
›Reveal solutionSolution
Tests the cell-constant method for molar conductivity: use the KCl calibration run to find the

cell constant, then apply it to the K $_2$ SO $_4$ run. Answer: $2.8\times10^{-2}\ \mathrm{S\,m^2\,mol^{-1}}$ .

Concept and Intuition

A conductivity cell has a fixed geometric cell constant $G^* = l/A$ (in $\mathrm{m^{-1}}$ ),

which relates the measured resistance to the solution's conductivity via

$\kappa = G^*/R$ (equivalently $G^* = \kappa R$ ). Once $G^*$ is found from a solution of known

conductivity (here, KCl), the same cell constant applies to any other solution placed in that

cell, letting us find the unknown solution's conductivity, and from that its molar conductivity

$\Lambda_m = \kappa/c$ (with $c$ expressed in $\mathrm{mol\,m^{-3}}$ to get SI units of $\Lambda_m$ ).

Step-by-Step Solution

Find the cell constant from the KCl calibration: $G^* = \kappa_{KCl} \times R_{KCl} = 0.28 \times 82.2 = 23.016\ \mathrm{m^{-1}}$ .

Use $G^*$ to find $\kappa$ of the K $_2$ SO $_4$ solution: $\kappa_{K_2SO_4} = G^*/R_{K_2SO_4} = 23.016/325 = 0.0708\ \mathrm{S\,m^{-1}}$ .

Convert concentration to SI (mol per m $^3$ ): $0.0025\ \mathrm{mol\,dm^{-3}} \times 1000 = 2.5\ \mathrm{mol\,m^{-3}}$ . …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$6\times10^{-3}\ \mathrm{ohm^{-1}cm^{-1}}$
›Reveal solutionSolution
Compute molarity of the BaCl $_2$ solution, then use $\Lambda_m = 1000\kappa/M$ to find molar conductivity. Answer: $x=1.2$ .

Concept and Intuition

Molar conductivity $\Lambda_m$ relates a solution's specific conductivity ( $\kappa$ ) to its molar concentration: $\Lambda_m = \dfrac{\kappa \times 1000}{M}$ (with $\kappa$ in $\mathrm{ohm^{-1}cm^{-1}}$ , $M$ in mol/L, giving $\Lambda_m$ in $\mathrm{ohm^{-1}cm^2mol^{-1}}$ ). We first need the molarity from the given mass and volume.

Step-by-Step Solution

Molar mass of $\mathrm{BaCl_2} = 137 + 2(35.5) = 208\ \mathrm{g/mol}$ .

Moles $= \dfrac{2.08}{208} = 0.01$ mol.

Volume $= 200\ \mathrm{mL} = 0.2$ L, so Molarity $= \dfrac{0.01}{0.2} = 0.05\ \mathrm{mol/L}$ . …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\Lambda_m^\circ$
›Reveal solutionSolution
Combine the three given limiting molar conductivities via Kohlrausch's law so that the $Ba^{2+}$ and $Cl^-$ contributions cancel, leaving $\Lambda_m^\circ(\mathrm{NH_4OH})$ .

Concept and Intuition

Kohlrausch's law states each ion contributes independently to the limiting molar conductivity, so ionic contributions can be added/subtracted across compounds that share ions. Here we want $\lambda^\circ(NH_4^+)+\lambda^\circ(OH^-)$ ; we can build it from $NH_4Cl$ (gives $NH_4^+ + Cl^-$ ) plus half of $Ba(OH)_2$ (gives $\tfrac12 Ba^{2+}+OH^-$ ) minus half of $BaCl_2$ (gives $\tfrac12 Ba^{2+}+Cl^-$ ), which cancels the $Ba^{2+}$ and $Cl^-$ contributions, leaving exactly $NH_4^++OH^-$ .

Step-by-Step Solution

Write ionic contributions: $\Lambda^\circ(Ba(OH)_2)=\lambda(Ba^{2+})+2\lambda(OH^-)=457.0$ ; $\Lambda^\circ(BaCl_2)=\lambda(Ba^{2+})+2\lambda(Cl^-)=240.6$ ; $\Lambda^\circ(NH_4Cl)=\lambda(NH_4^+)+\lambda(Cl^-)=2130$ .

Subtract the first two and halve: $\lambda(OH^-)-\lambda(Cl^-) = \dfrac{457.0-240.6}{2}=108.2$ . …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$\mathrm{S\,cm^2\,mol^{-1}}$
›Reveal solutionSolution
Using $\alpha=\Lambda_m/\Lambda_m^\circ$ and Ostwald's dilution law $K_a=\dfrac{c\alpha^2}{1-\alpha}$ gives $K_a=3.0\times10^{-4}$ for this weak acid.

Concept and Intuition

For a weak electrolyte, the fraction of molar conductivity actually achieved at a finite concentration compared to its limiting (infinite-dilution) value directly gives the degree of dissociation $\alpha$ (this is Arrhenius's conductance-based definition of $\alpha$ ). Once $\alpha$ is known, Ostwald's dilution law connects it to the acid's dissociation constant.

Step-by-Step Solution

Limiting molar conductivity: $\Lambda_m^\circ = \lambda^\circ_{H^+} + \lambda^\circ_{HCOO^-} = 349.6 + 54.6 = 404.2\,\text{S cm}^2\text{mol}^{-1}$ .

Degree of dissociation: $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{40.42}{404.2} = 0.1$ .

Ostwald's dilution law: $K_a = \dfrac{c\,\alpha^2}{1-\alpha}$ .

Substitute $c=0.027\,\text{M}$ , $\alpha=0.1$ : $K_a = \dfrac{0.027\times(0.1)^2}{1-0.1} = \dfrac{0.027\times0.01}{0.9} = \dfrac{2.7\times10^{-4}}{0.9} = 3.0\times10^{-4}$ .

Common Mistakes …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$CH_3CO_2H$
›Reveal solutionSolution
Combining the given limiting molar conductivities of $(CH_3COO)_2Ba$ , $BaCl_2$ , and $HCl$ via Kohlrausch's law of independent ion migration reconstructs $\Lambda^0_m(CH_3COOH)$ as $(x_1-x_2)/2+x_3$ .

Concept and Intuition

Kohlrausch's law says the limiting molar conductivity of any electrolyte is the sum of the limiting ionic conductivities of its constituent ions, each independent of the other ion present. This lets us build the conductivity of a weak electrolyte (like acetic acid) from strong electrolytes sharing common ions.

Step-by-Step Solution

Write each given compound in terms of ionic conductivities: $x_1 = \Lambda^0_m((CH_3COO)_2Ba) = \lambda^0(Ba^{2+}) + 2\lambda^0(CH_3COO^-)$ $x_2 = \Lambda^0_m(BaCl_2) = \lambda^0(Ba^{2+}) + 2\lambda^0(Cl^-)$ $x_3 = \Lambda^0_m(HCl) = \lambda^0(H^+) + \lambda^0(Cl^-)$

Subtract: $x_1 - x_2 = 2\lambda^0(CH_3COO^-) - 2\lambda^0(Cl^-)$ , so $\dfrac{x_1-x_2}{2} = \lambda^0(CH_3COO^-) - \lambda^0(Cl^-)$ . …
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Q.Match the terms given in Column I with the units given in Column II. Column I:

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Q.Match the terms given in Column I with the units given in Column II.
Column I:

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