Q.The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146×10−3 S cm−1?
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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 …
The key idea is that cell constant G∗ relates measured resistance R to the solution's conductivity κ by the formula:
G∗=κ×R
Step 1: Write the relation between conductivity, resistance, and cell constant.
κ=RG∗⇒G∗=κ×R …
The cell constant is the product of measured resistance and known conductivity. Using G∗=κ×R, we get G∗=0.146×10−3×1500=0.219 cm−1.
The key idea here is simple: a conductivity cell has a fixed geometry — the distance between electrodes divided by their area — which we call the cell constant (G∗). You can’t measure it directly with a ruler inside the solution, but you can find it electrically.
Conductivity (κ) is an intrinsic property of the solution. Resistance (R) is what you measure. They are related by:
κ=R1×G∗
Rearranging:
G∗=κ×R
That’s the whole backbone. Once you know the conductivity of a standard solution (here, 0.001 M KCl at 298 K, whose conductivity is tabulated and given), and you measure its resistance in your cell, the cell constant follows immediately.
Let’s walk through it.
-
Write down what’s given.
Resistance, R=1500 Ω
Conductivity, κ=0.146×10−3 S cm−1
(Notice the units: Siemens per centimetre — that’s the clue that our cell constant will come out in cm−1.)
-
Apply the relation.
G∗=κ×R=(0.146×10−3)×1500
- Do the arithmetic.
0.146×10−3=1.46×10−4
Multiply by 1500:
1.46×10−4×1500=1.46×10−4×1.5×103=1.46×1.5×10−1=2.19×10−1=0.219
- Attach the unit. …
Method: Cell Constant from Conductivity and Resistance
This problem uses the fundamental relationship between conductivity (κ), resistance (R), and the cell constant (G∗).
Concept First
The cell constant G∗ (also written as l/A) is a fixed property of a conductivity cell — it depends only on the geometry (distance between electrodes divided by their area). It links the measured resistance R to the conductivity κ of the solution:
κ=G∗×R1
Rearranging:
G∗=κ×R
Steps
-
Identify given data
- Conductivity, κ=0.146×10−3 S cm−1
- Resistance, R=1500 Ω
-
Apply the formula
G∗=κ×R
- Substitute and calculate
G∗=(0.146×10−3)×1500
G∗=0.146×1.5
G∗=0.219 cm−1 …
Common Mistakes & How to Avoid Them
Mistake 1: Inverting the Cell-Constant Formula
The error: Students write G∗=κR instead of G∗=κ×R, or confuse which quantity is measured and which is looked up.
Why it happens: Conductivity (κ) is related to conductance (G=1/R) by κ=G×G∗, so it's easy to rearrange incorrectly under time pressure.
How to avoid:
- Start from the definition: κ=R1×G∗
- Rearranging for the cell constant: G∗=κ×R
- Sanity check: cell constant is a fixed geometric number (typically 0.1–10 cm−1 for standard cells) — if your answer looks like a resistance or a conductivity, you've used the wrong formula.
Mistake 2: Mishandling the 10−3 in κ=0.146×10−3 S cm−1
The error: Students drop the ×10−3 or misplace the decimal when multiplying by R=1500 Ω, landing on 219 cm−1 or 0.0219 cm−1 instead of 0.219 cm−1.
How to avoid:
- Convert to plain decimal or keep scientific notation consistently: κ=0.146×10−3=1.46×10−4 S cm−1
- Multiply: G∗=1.46×10−4×1500=1.46×10−4×1.5×103=2.19×10−1=0.219 cm−1
Mistake 3: Forgetting Cell Constant Has Units of cm−1
The error: Students report the answer as a pure number (e.g., "0.219") with no unit, or attach the wrong unit like S cm−1 (the unit of κ, not G∗).
Why it happens: G∗=κ×R, and since κ is S cm−1 and R is Ω=S−1, the Siemens cancel, leaving only cm−1 — students often don't carry the units through the multiplication to see this.
How to avoid:
- Track units explicitly: S cm−1×Ω=S cm−1×S−1=cm−1
- Always write the final answer as G∗=0.219 cm−1, never as a bare number.
Mistake 4: Confusing the Standard KCl Solution With an Unknown
The error: Students think the 0.001 M KCl and its conductivity are the unknown being solved for, rather than realising this is a standard reference solution used only to calibrate the cell.
How to avoid: …
- KEAM 2026Set eng-2026-04214 marksMCQQ.The molar conductivity of a weak mono basic acid, HA at 298 K is 70 Scm2mol−1. What is the percentage ionisation of HA at 298 K? [At infinite dilution λH+=340 Scm2mol−1 and λA−=80 Scm2mol−1] (A) 8.35 % (B) 16.7 % (C) 20 % (D) 32.5 % (E) 15.3 %
›Reveal solutionSolution
α=Λm/Λm∘=70/420=16.7.
Limiting molar conductivity of HA:
Λm∘=λH+∘+λA−∘=340+80=420,Scm2mol−1.
Degree of ionisation: …
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.The limiting molar conductance for aqueous solution of CaCl2 at 298K is 271.6 S cm2 mol−1. If the limiting ionic conductance of Ca2+ ion at the same temperature is 119 S cm2 mol−1 what is the limiting ionic conductance of Cl− ion? (A) 152.6 S cm2 mol−1 (B) 76.3 S cm2 mol−1 (C) 135.8 S cm2 mol−1 (D) 228.7 S cm2 mol−1 (E) 114.35 S cm2 mol−1
›Reveal solutionSolution
Kohlrausch: Λm∘(CaCl2)=λ∘(Ca2+)+2λ∘(Cl−).
271.6=119+2λ∘(Cl−) …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The molar conductivity (Λm) acetic acid is 78.1 S cm2 mol−1. Its degree of dissociation (α) is (Λm0) for acetic acid = 390.5 S cm2 mol−1 (A) 0.12 (B) 0.40 (C) 0.02 (D) 0.20 (E) 0.04
›Reveal solutionSolution
Degree of dissociation of a weak electrolyte α=Λm0Λm=390.578.1=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:
α=Λm0Λm …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.The ion with the highest limiting molar conductance at 298 K is (A) H+ (B) Na+ (C) K+ (D) Ca2+ (E) Mg2+
›Reveal solutionSolution
The proton conducts by the Grotthuss (hopping) mechanism, giving H+ by far the largest limiting molar conductivity (≈350 S cm2 mol−1).
Limiting molar conductivities at 298 K (in S cm2 mol−1) are roughly:
- H+≈349.8
- K+≈73.5
- Ca2+≈119 (but per mole of charge ≈59.5)
- Na+≈50.1
- Mg2+≈106 (per mole of charge ≈53) …
- KEAM 2024Set eng-2024-06074 marksMCQQ.The conductivity of 0.02 mol L−1 KCl solution is 0.248 S m−1. Its molar conductivity is (A) 20 S m2 mol−1 (B) 1.24×10−3 S m2 mol−1 (C) 1.24×10−4 S m2 mol−1 (D) 2.48×10−2 S m2 mol−1 (E) 1.24×10−2 S m2 mol−1
›Reveal solutionSolution
Molar conductivity Λm=κ/c. With κ=0.248 S m−1 and c=0.02 mol L−1=20 mol m−3, Λm=1.24×10−2 S m2 mol−1.
Molar conductivity is related to conductivity by
Λm=cκ.
Using SI units, the concentration must be in mol m−3:
c=0.02 mol L−1=0.02×1000=20 mol m−3. …
- KEAM 2024Set eng-2024-06084 marksMCQQ.The limiting molar conductances of NaCl, HCl and CH3COONa at 300 K are 126.4, 425.9 and 91.0 S cm2 mol−1 respectively. The limiting molar conductance of acetic acid at 300 K is (A) 266 S cm2 mol−1 (B) 390.5 S cm2 mol−1 (C) 461.3 S cm2 mol−1 (D) 208 S cm2 mol−1 (E) 108 S cm2 mol−1
›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 cm2 mol−1.
Acetic acid dissociates into CH3COO− and H+. Combining known limiting conductances:
Λm∘(CH3COOH)=Λm∘(CH3COONa)+Λm∘(HCl)−Λm∘(NaCl). …
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.The SI unit of molar conductivity is (A) S m3 mol−1 (B) S m mol−1 (C) S m mol−2 (D) S m2 mol−1 (E) S m2 mol−2
›Reveal solutionSolution
The SI unit of molar conductivity is S m2 mol−1.
Concept and Intuition
Molar conductivity is the conductivity per unit molar concentration, defined as Λm=κ/c. Its unit follows directly from the units of conductivity and concentration in SI.
Step-by-Step Solution
- Conductivity κ has SI unit S m−1.
- Concentration c in SI is mol m−3. …
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