Q.The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm−1. Calculate its molar conductivity.
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🔒 Start your 14-day free trial to unlock the full solution →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.
Conductance (G) is the reciprocal of resistance:
G=R1
Unit: siemens (S) — named after Werner von Siemens. 1 S = 1 A/V (ampere per volt).
If a wire has R=10 Ω, its conductance is G=0.1 S. If R=0.5 Ω, G=2 S — it conducts twice as well.
Ohm's Law in Conductance Form
You know V=IR. Rearranging:
I=RV=GV
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" ρ) and the geometry (length L, cross-sectional area A):
R=ρAL
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 (σ):
σ=ρ1
And for a uniform wire:
G=σLA
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 σ → good conductor (copper: ≈5.8×107 S/m).
- Low σ → poor conductor / insulator (glass: ≈10−12 S/m).
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 | Ω |
| Resistivity | ρ | RA/L | Material only | Ω⋅m |
| Conductance | G | 1/R | Material + geometry | S |
| Conductivity | σ | 1/ρ | 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 (σ).
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=RV
Define conductance G as the reciprocal of resistance:
G=R1
So:
I=GV
Why this makes sense:
- A larger G means more current for the same voltage — the conductor "conducts" better.
- G has units of siemens (S) = 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 ρ (resistivity)
The formula:
R=ρAL
Now, conductivity σ is the reciprocal of resistivity:
σ=ρ1
So:
R=σ1⋅AL
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=σLA
This is the central relationship. Let's see why it holds:
- σ 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 σ (like copper) gives higher G than one with low σ (like iron), for the same shape.
5. Microscopic Derivation (Why σ Exists)
At the microscopic level, conductivity arises from electron motion:
σ=neμ
Where:
- n = number of free electrons per unit volume
- e = electron charge …
Concept: Molar Conductivity — the conductivity of all ions produced from one mole of electrolyte in solution.
The formula is:
Λm=Cκ×1000
where κ is conductivity in S cm−1, C is molarity in mol L−1, and the factor 1000 converts cm3 to L.
Step 1: Identify the given values.
κ=0.0248 S cm−1, C=0.20 mol L−1.
Step 2: Substitute into the formula. …
Molar conductivity is the conductivity of a solution containing one mole of electrolyte, calculated as Λm=cκ. For this 0.20 M KCl solution, Λm=124 S cm2 mol−1.
Molar conductivity (Λm) is a way to compare how well different electrolytes conduct electricity, independent of their concentration. Think of it this way: conductivity (κ) tells you how much current flows through a given volume of solution. But if you have a more concentrated solution, there are simply more ions present to carry charge — so the conductivity goes up just because there are more carriers. Molar conductivity removes this "crowding" effect by normalising to a fixed amount of electrolyte (one mole). It answers: If I had exactly one mole of this electrolyte dissolved, how well would it conduct?
The formula is straightforward:
Λm=cκ
where κ is the measured conductivity (in S cm−1) and c is the molar concentration (in mol L−1). The tricky part — and the reason many students slip up — is the units. Conductivity is usually given in S cm−1, but concentration is in mol L−1. One litre is 1000 cm3, so you must convert the concentration to mol cm−3 before dividing, or equivalently, multiply by 1000 after dividing. Let's walk through it.
-
Write down what's given.
κ=0.0248 S cm−1
c=0.20 M=0.20 mol L−1
-
Convert concentration to mol cm−3.
Since 1 L=1000 cm3,
c=1000 cm30.20 mol=2.0×10−4 mol cm−3
- Apply the formula.
Λm=cκ=2.0×10−4 mol cm−30.0248 S cm−1
Dividing:
Λm=2.0×10−40.0248 S cm2 mol−1=124 S cm2 mol−1 …
Method: Direct Formula Substitution
This is the most straightforward method — simply apply the definition of molar conductivity.
Concept (Why this works)
Molar conductivity (Λm) tells us the conducting power of all ions produced from one mole of an electrolyte. It relates the measured conductivity (κ) to the concentration (c) of the solution.
Formula
Λm=cκ×1000
Where:
- κ = conductivity in S cm⁻¹
- c = concentration in mol L⁻¹
- Factor 1000 converts cm³ to L (since 1 L = 1000 cm³)
Steps
-
Identify given data
- κ=0.0248 S cm−1
- c=0.20 M=0.20 mol L−1
-
Apply the formula
Λm=0.200.0248×1000
-
Simplify
Λm=0.2024.8=124 …
Here are the common mistakes students make when calculating molar conductivity from conductivity data, along with how to avoid each.
1. Forgetting to Convert Units (The Most Common Mistake)
The Mistake:
Students plug the given conductivity (0.0248 S cm−1) directly into the formula without checking units. The molar conductivity formula requires conductivity in S cm⁻¹ and concentration in mol cm⁻³, but concentration is often given in mol L⁻¹ (or M).
How to Avoid:
Always write the formula first and check every unit.
- Molar conductivity:
Λm=cκ
where κ is conductivity (S cm⁻¹) and c is concentration (mol cm⁻³).
- Given: c=0.20 M=0.20 mol L−1 Convert to mol cm⁻³:
1 L=1000 cm3⇒c=10000.20=2.0×10−4 mol cm−3
- Then:
Λm=2.0×10−40.0248=124 S cm2 mol−1
Key takeaway: Always convert M → mol cm⁻³ by dividing by 1000.
2. Using the Wrong Formula or Confusing Conductivity with Molar Conductivity
The Mistake:
Some students use Λm=κ×c (multiplying instead of dividing) or confuse κ (conductivity) with Λm (molar conductivity).
How to Avoid:
Remember the definition: Molar conductivity is the conductivity of a solution containing 1 mole of electrolyte placed between electrodes 1 cm apart.
- If conductivity is high for a dilute solution, molar conductivity is large.
- The relationship is inverse: Λm=cκ.
Mnemonic: “Conductivity per mole” → divide by concentration.
3. Ignoring the Temperature Dependence
The Mistake:
Students assume the same formula works at any temperature without noting that conductivity changes with temperature. The problem explicitly states 298 K — using a different temperature’s data or formula would be wrong.
How to Avoid:
Always note the temperature given. For exam problems, use the data as provided. If temperature is not given, assume standard conditions (298 K). Never mix data from different temperatures.
4. Misplacing Decimal Points in Unit Conversion
The Mistake:
When converting 0.20 M to mol cm⁻³, students sometimes write 0.20/1000=0.00020 but then misplace the decimal in the final division, e.g., 0.0248/0.00020=1240 instead of 124.
How to Avoid:
Use scientific notation for clarity:
- c=2.0×10−4 mol cm−3
- κ=2.48×10−2 S cm−1
- Then: …
Showing the 12 most recent of 32 on this concept.
- CBSE 2025Set ANNUAL1 markQ.What is the SI unit of molar conductivity?
›Reveal solutionSolution
Molar conductivity's SI unit is S m² mol⁻¹.
Molar conductivity Λm=Cκ, where κ (conductivity) has SI unit Sm−1 and concentration C has SI unit molm−3.
Λm=molm−3Sm−1=Sm2mol−1
…
- CBSE 2025Set ANNUAL1 markQ.Define the following — Limiting molar conductivity
›Reveal solutionSolution
Limiting molar conductivity is molar conductivity extrapolated to zero concentration.
Limiting molar conductivity (Λm0 or Λm∞) 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, Λm0 is obtained by extrapolating the Λm …
- CBSE 2025Set ANNUAL1 markMCQQ.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
…
- CBSE 2025Set ANNUAL1 markQ.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. …
- CBSE 2025Set ANNUAL1 markMCQQ.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:
Λm=Cκ×1000
where κ (specific conductivity) has units S cm^-1 and C (concentration) has units mol L^-1 (mol per 1000 cm^3).
…
- CBSE 2025Set ANNUAL1 markMCQQ.The molar conductivity of a 0.1mol L−1 solution of KCl with electrolytic conductivity 0.0129 S cm−1 at 298 K is –(a) 12.9 S cm2 mol−1(b) 1.29 S cm2 mol−1(c) 0.0129 S cm2 mol−1(d) 129 S cm2 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, κ) and molar concentration C (in molL−1) by:
Λm=Cκ×1000
…
- CBSE 2025Set ANNUAL1 markMCQQ.The formula used to calculate molar conductivity of an electrolyte is _____.(a) Λ=k1000c(b) c=k1000Λ(c) Λ=c1000k(d) k=Λc1000
›Reveal solutionSolution
Λm=c1000κ.
Molar conductivity (Λm) relates to specific conductivity (κ) and molar concentration c (in moldm−3) by:
Λm=cκ×1000
…
- CBSE 2025Set ANNUAL1 markQ.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 Λm=κ/C: siemens metre-squared per mole.
Molar conductivity is defined as Λm=Cκ, where κ (specific/electrical conductivity) has SI unit Sm−1 and C (molar concentration) has SI unit molm−3. Dividing, the SI unit of Λm works out to
molm−3Sm−1=Sm2mol−1.
(In practical lab work, where κ is often expressed in Scm−1 and C in molL−1, the commonly used relation is Λm=C1000κ, giving the c.g.s.-style unit Scm2mol−1.)
…
- CBSE 2024Set A11 markMCQQ.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 Λm 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), Λm reaches a limiting maximum value denoted Λm∘, the **limiting molar co …
- CBSE 2024Set B1 markQ.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, Λeq, is defined as Λeq=κ×V, where κ (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 …
- CBSE 2024Set ANNUAL1 markQ.The molar conductivity of 2.5×10−2 M of methanoic acid is 46.1 S cm2 mol−1. Calculate the degree of dissociation. (Molar conductivity of H+ and HCOO− at infinite dilutions are λ°H+=349.6 S cm2 mol−1 and λ°HCOO−=54.6 S cm2 mol−1).
›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:
α=Λm°Λm
Limiting molar conductivity of methanoic acid (using Kohlrausch's law of independent migration of ions):
Λm°=λ°H++λ°HCOO−=349.6+54.6=404.2 Scm2mol−1 …
- CBSE 2024Set ANNUAL1 markQ.Write SI unit of molar conductivity.
›Reveal solutionSolution
SI unit of molar conductivity is Sm2mol−1.
Molar conductivity Λm 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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