Q.Resistance of a conductivity cell filled with 0.1 mol L−1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L−1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L−1 KCl solution. The conductivity of 0.1 mol L−1 KCl solution is 1.29 S m−1.
Concept understanding — Conductance And 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).
- Resistivity = the "roughness" of the pipe's inner surface (material property).
- Conductivity = the "smoothness" of the pipe's inner surface (material property).
A copper pipe is smooth (high conductivity). A rubber hose is rough (low conductivity). But a short, fat rubber hose might still have decent conductance — because geometry can compensate for poor material.
Key Takeaway for Exams
- G=R1 and σ=ρ1.
- G=σLA for a uniform conductor.
- Conductivity is an intrinsic material property; conductance is an extrinsic property of a specific object.
- In circuits, you'll often use conductance when dealing with parallel resistors (total conductance = sum of individual conductances).
You now have the complete picture: from resistance to conductance, from resistivity to conductivity — and the clean relationship between them.
Conductance and conductivity are core quantitative ideas in the NCERT/CBSE Class 12 Chemistry Electrochemistry chapter, and ‘conductance vs conductivity formula’ is a regularly asked important question in board exams as well as JEE Main and NEET chemistry sections. This relationship also feeds directly into later topics like molar conductivity and Kohlrausch's law.
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
- μ = electron mobility (how fast they drift per unit electric field)
Why this works:
- More free electrons (n large) → more charge carriers → higher conductivity.
- Higher mobility (μ large) → electrons move faster for the same push → higher conductivity.
This explains why metals (high n) are good conductors, and why heating reduces σ (more collisions → lower μ).
6. Summary: The Logical Chain
| Step | Concept | Formula | Why |
|---|---|---|---|
| 1 | Ohm's law | V=IR | Voltage drives current against resistance |
| 2 | Conductance | G=1/R | Measures ease of flow |
| 3 | Resistivity | R=ρL/A | Geometry + material |
| 4 | Conductivity | σ=1/ρ | Material's intrinsic ability |
| 5 | Key result | G=σLA | Combines material + geometry |
Final takeaway:
Conductance G is not just a number — it's the product of how good the material is (σ) and how the shape helps (A/L). This is why a thick copper wire conducts far better than a thin iron wire of the same length.
Concept: Conductance and Conductivity — the cell constant G∗ links measured resistance R to conductivity κ via κ=G∗/R.
Step 1: Find the cell constant
For the 0.1 mol L−1 solution:
κ1=1.29 S m−1, R1=100 Ω
G∗=κ1×R1=1.29×100=129 m−1
Step 2: Conductivity of 0.02 mol L−1 solution
R2=520 Ω
κ2=R2G∗=520129=0.2481 S m−1
Step 3: Molar conductivity
Concentration c=0.02 mol L−1=20 mol m−3
Λm=cκ2=200.2481=0.012405 S m2 mol−1
The conductivity is 0.248 S m−1 and the molar conductivity is 1.24×10−2 S m2 mol−1.
The cell constant is found from the known conductivity and resistance of the 0.1 M KCl solution. Using that constant, the conductivity of the 0.02 M KCl solution is calculated from its resistance. Molar conductivity is then obtained by dividing conductivity by concentration (in mol/m³). The final values are κ=0.248 S m−1 and Λm=1.24×10−2 S m2 mol−1.
Why this works: Conductance and conductivity
The key idea is that a conductivity cell has a fixed geometry — the distance between electrodes and their area don't change. This geometry is captured by the cell constant, G∗, with units of m−1:
G∗=area of electrodesdistance between electrodes=Al
Conductance G (in siemens, S) is the reciprocal of resistance: G=1/R. Conductivity κ (in S m−1) is related to conductance by:
κ=G×G∗=RG∗
So if we know κ and R for one solution, we can find G∗. Then for any other solution in the same cell, knowing R gives κ.
Molar conductivity Λm (in S m2 mol−1) is then:
Λm=cκ
where c is concentration in mol m−3. This is the conductivity per mole of electrolyte — it tells us how well each mole carries current.
Step-by-step solution
1. Find the cell constant using the 0.1 M KCl data.
We are given:
- For 0.1 mol L−1 KCl: R1=100 Ω, κ1=1.29 S m−1
From κ=G∗/R, we get:
G∗=κ1×R1=1.29 S m−1×100 Ω=129 m−1
Notice the units: S m−1×Ω=m−1 because siemens is the reciprocal of ohm (S=Ω−1). So the cell constant comes out in m−1, as expected.
2. Calculate the conductivity of the 0.02 M KCl solution.
For the same cell, G∗ is fixed. With R2=520 Ω:
κ2=R2G∗=520 Ω129 m−1=0.248 S m−1
A common mistake is to forget that resistance is in ohms and cell constant in m−1, giving conductivity in S m−1 directly. But if you used cm instead of m, you'd be off by a factor of 100. Always check units: here everything is in SI.
3. Convert concentration to SI units (mol m−3).
The given concentration is 0.02 mol L−1. Since 1 L=10−3 m3:
c=0.02 mol L−1=0.02×103 mol m−3=20 mol m−3
4. Compute molar conductivity.
Λm=cκ2=20 mol m−30.248 S m−1=0.0124 S m2 mol−1
In scientific notation:
Λm=1.24×10−2 S m2 mol−1
Molar conductivity is often expressed in S cm2 mol−1 in some textbooks. To convert: 1 S m2 mol−1=104 S cm2 mol−1, so here it would be 124 S cm2 mol−1. But since the problem gave conductivity in S m−1, we stick with SI.
The conductivity of 0.02 mol L−1 KCl is 0.248 S m−1 and its molar conductivity is 1.24×10−2 S m2 mol−1.
Method: Cell Constant Method
This method uses the cell constant (G∗) of the conductivity cell, which remains fixed for a given cell. The cell constant relates resistance (R) to conductivity (κ) via:
κ=RG∗
Step 1: Find the cell constant using the known solution
For the 0.1 mol L−1 KCl solution:
- Given: R1=100 Ω, κ1=1.29 S m−1
Using the formula:
G∗=κ1×R1
G∗=1.29×100=129 m−1
Cell constant G∗=129 m−1
Step 2: Calculate conductivity of the 0.02 mol L−1 KCl solution
For the unknown solution:
- Given: R2=520 Ω
- Cell constant is the same: G∗=129 m−1
κ2=R2G∗=520129
κ2=0.248 S m−1
Conductivity κ2=0.248 S m−1
Step 3: Calculate molar conductivity of the 0.02 mol L−1 KCl solution
Molar conductivity (Λm) is given by:
Λm=cκ
where:
- κ is in S m−1
- c is concentration in mol m−3
Convert concentration:
0.02 mol L−1=0.02×1000=20 mol m−3
Now:
Λm=200.248=0.0124 S m2 mol−1
Molar conductivity Λm=1.24×10−2 S m2 mol−1
Final Answer
| Quantity | Value |
|---|---|
| Conductivity (κ) | 0.248 S m−1 |
| Molar conductivity (Λm) | 1.24×10−2 S m2 mol−1 |
Here are the most common mistakes students make with this problem, along with the conceptual fixes to avoid them.
1. Forgetting the Cell Constant is the Bridge
The Mistake: Students try to directly use the formula κ=R1×Al without first calculating the cell constant (G∗=l/A) from the known data.
Why it happens: They see two resistances and two concentrations and panic, trying to plug numbers into the wrong formula.
How to Avoid:
- Concept: The cell constant (l/A) is a property of the physical cell (the distance between electrodes and their area). It does not change when you change the solution.
- Action: Always calculate G∗ first using the data for the known solution (0.1 mol L−1 KCl).
G∗=κ×R
G∗=(1.29 S m−1)×(100 Ω)=129 m−1
2. Unit Confusion (cm vs. m)
The Mistake: Using κ in S cm−1 when the problem gives κ in S m−1, or forgetting to convert concentration from mol L−1 to mol m−3 for molar conductivity.
Why it happens: Electrochemistry problems often mix units. Conductivity is often given in S cm−1 in textbooks, but here it's in S m−1.
How to Avoid:
- Check units at the start. The given κ=1.29 S m−1.
- Cell constant will be in m−1 (since R is in Ω).
- Conductivity of the unknown (κ0.02) will come out in S m−1.
- For molar conductivity (Λm): Convert concentration from mol L−1 to mol m−3.
0.02 mol L−1=0.02×1000=20 mol m−3
3. Using the Wrong Resistance for the Cell Constant
The Mistake: Using the resistance of the 0.02 mol L−1 solution (520 Ω) to calculate the cell constant.
Why it happens: Students think "cell constant" is calculated from the solution they are trying to find, not from the standard/reference solution.
How to Avoid:
- Rule: The cell constant is always calculated from the solution whose conductivity is known.
- Here, the known solution is 0.1 mol L−1 KCl with κ=1.29 S m−1 and R=100 Ω.
- The 0.02 mol L−1 solution is the unknown — you use its resistance after you have G∗.
4. Confusing Conductivity (κ) with Conductance (G)
The Mistake: Thinking that 1/R (conductance) is the same as conductivity (κ).
Why it happens: The words sound similar, and both involve resistance.
How to Avoid:
- Remember the relationship:
κ=G×(Al)
where $G = 1/R$.
- Conductivity (κ) is conductance per unit length and area — it's an intensive property of the solution.
- Conductance (G) depends on the geometry of the cell.
5. Forgetting the Final Step: Molar Conductivity
The Mistake: Stopping after finding conductivity (κ) and not calculating molar conductivity (Λm).
Why it happens: The question explicitly asks for both conductivity and molar conductivity, but students rush.
How to Avoid:
- Read the question twice. Underline "conductivity and molar conductivity".
- Formula:
Λm=cκ
where $c$ is in $\text{mol m}^{-3}$.
- Plug in carefully:
Λm=20 mol m−30.248 S m−1=0.0124 S m2mol−1
Quick Summary Checklist
| Step | Common Mistake | Correct Approach |
|---|---|---|
| 1. Cell Constant | Use R of unknown solution | Use R and κ of known solution |
| 2. Units | Mix cm and m | Keep everything in meters and S m−1 |
| 3. Conductivity | Confuse G and κ | κ=G∗/R |
| 4. Molar Conductivity | Forget to convert L to m3 | Multiply concentration by 1000 |
| 5. Final Answer | Stop at κ | Calculate Λm too |
Final Correct Values (for your reference):
- Cell constant: 129 m−1
- Conductivity of 0.02 M KCl: 0.248 S m−1
- Molar conductivity: 1.24×10−2 S m2mol−1
Showing the 12 most recent of 27 on this concept.
- CBSE 2026Set A1 markMCQQ.On increasing dilution, the specific conductance of an electrolyte(a) increases(b) decreases(c) remains constant(d) none of these
›Reveal solutionSolution
Specific conductance (conductance per unit volume) falls on dilution because the number of current-carrying ions per unit volume decreases.
Specific conductance (κ) is the conductance of a solution held between electrodes 1 cm apart with 1 cm² area, i.e. conductance of unit volume. On dilution the number of ions per unit volume decreases, so κ decreases. (In contrast, molar conductance Λm increases on dilution because it accounts for all ions from one mole.)
✓Final answer(b) decreases.
- CBSE 2026Set A1 markMCQQ.The number of ions in aqueous solution of [Co(NH3)5Cl]Cl2 is(a) 3(b) 4(c) 2(d) 6
›Reveal solutionSolution
Only the ions outside the coordination sphere are free; [Co(NH3)5Cl]Cl2 gives one complex cation plus two chloride ions = 3 ions.
In a coordination compound, only the counter ions outside the square brackets dissociate in water; the ligands inside the coordination sphere stay bound to the metal. Here one Cl and five NH3 are coordinated to cobalt, and two Cl are counter ions:
[Co(NH3)5Cl]Cl2 -> [Co(NH3)5Cl]2+ + 2 Cl-
That gives 1 complex cation + 2 chloride anions = 3 ions in solution. (Only the two ionisable chlorides would be precipitated by AgNO3.)
✓Final answer(a) 3 — one complex cation and two chloride ions.
- CBSE 2026Set ANNUAL1 markMCQQ.The unit of cell constant is:(a) Ohm^-1 cm^2(b) cm^-1(c) Ohm^-1 cm^-1(d) Ohm^-1 cm^2/ g eq
›Reveal solutionSolution
Cell constant G∗=l/A has the unit of reciprocal length, i.e. cm^-1.
The cell constant of a conductivity cell is defined as the ratio of the distance between the two electrodes (l) to the area of cross-section of the electrodes (A): G∗=Al. Since l has units of cm and A has units of cm^2, the cell constant has units of cm2cm=cm−1.
This is distinct from conductivity κ (specific conductance), whose unit is Ω−1cm−1 (obtained as G∗× the measured conductance 1/R, unit Ω−1), and from molar conductivity Λm, whose unit is Ω−1cm2mol−1. The cell constant itself is purely a geometric quantity of the cell, hence its unit is just cm^-1.
✓Final answer(b) cm^-1 is the correct unit of the cell constant.
- CBSE 2026Set ANNUAL1 markMCQQ.The unit of specific conductivity is:(a) ohm⁻¹(b) ohm⁻¹ cm⁻¹(c) ohm cm(d) ohm cm⁻¹
›Reveal solutionSolution
Specific conductance (κ) is measured in ohm⁻¹ cm⁻¹ (S cm⁻¹).
Specific conductivity (κ), also called conductivity, is the conductance of a 1 cm cube of a solution of an electrolyte. Conductance (G) is the reciprocal of resistance and is measured in ohm⁻¹ (siemens, S). Since κ=G×(l/A), where l/A (the cell constant) has units of cm⁻¹, the derived unit of specific conductivity works out to ohm⁻¹ cm⁻¹, i.e. S cm⁻¹.
✓Final answer(b) ohm⁻¹ cm⁻¹.
- CBSE 2025Set 56/5/11 markMCQQ.Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : For measuring resistance of an ionic solution an AC source is used. Reason (R) : Concentration of ionic solution will change if DC source is used.
›Reveal solutionSolution
AC is used to measure ionic solution resistance because DC causes electrolysis, which changes the solution's composition and hence its conductance; both statements are true and the reason correctly explains the assertion.
When we measure the resistance (or conductance) of an ionic solution, we're essentially probing how easily ions can carry current through the liquid. The choice between AC and DC isn't arbitrary—it stems from what happens at the electrode-solution interface.
Why DC causes problems
In an ionic solution, current flows via the movement of ions: cations migrate toward the cathode, anions toward the anode. With a DC source, these ions don't just move—they undergo redox reactions at the electrodes. For instance, in a NaCl solution, Cl− ions get oxidized at the anode (2Cl−→Cl2+2e−) and H+ from water gets reduced at the cathode (2H++2e−→H2). This is electrolysis.
The consequence? The concentration of ions in the solution changes continuously. As ions are consumed or new species are produced, the conductance of the solution drifts. You're no longer measuring the property of the original solution—you're measuring a changing system. The reading becomes unreliable and time-dependent.
Why AC solves this
An alternating current reverses direction many times per second (typically at 1000 Hz or so in conductivity bridges). In one half-cycle, a tiny bit of electrolysis might begin, but in the next half-cycle the current reverses and the reaction is essentially undone. The net chemical change over many cycles is negligible. The solution composition remains stable, and the resistance measurement reflects the true, steady-state property of the ionic solution.
Watch outEven with AC, if the frequency is too low or the voltage too high, some net electrolysis can occur. Standard conductivity meters use optimized frequency and amplitude to minimize this.
Evaluating the statements
-
Assertion (A): "For measuring resistance of an ionic solution an AC source is used."
This is true. AC is the standard choice precisely to avoid the complications of electrolysis.
-
Reason (R): "Concentration of ionic solution will change if DC source is used."
This is also true. DC drives continuous electrolysis, altering ion concentrations.
-
Does R explain A?
Yes, it does. The reason we use AC (Assertion) is exactly because DC changes the concentration (Reason), which would invalidate the measurement. The reason is the correct explanation of the assertion.
✓Final answerThe correct option is (A): Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.
-
- CBSE 2025Set D1 markMCQQ.The unit of specific conductance is(a) ohm cm^-1(b) ohm cm^-2(c) ohm^-1 cm^-1(d) ohm^-1 cm^-2
›Reveal solutionSolution
Specific conductance = 1/(specific resistance), so its unit is ohm^-1 cm^-1 (S cm^-1).
Specific conductance (conductivity), kappa, is the reciprocal of specific resistance (resistivity), rho:
kappa = 1/rho
Specific resistance has the unit ohm cm, so its reciprocal has the unit:
kappa = 1 / (ohm cm) = ohm^-1 cm^-1 = S cm^-1
Thus the unit of specific conductance is ohm^-1 cm^-1.
✓Final answer(C) ohm^-1 cm^-1.
- CBSE 2025Set A1 markQ.Write the value of conductivity of superconductor.
›Reveal solutionSolution
Since conductivity is the reciprocal of resistivity, and a superconductor's resistivity drops to exactly zero, its conductivity becomes infinite.
Certain materials, when cooled below a characteristic critical temperature, lose all electrical resistance completely — this state is called superconductivity, and such materials are superconductors. Electrical conductivity (κ) and resistivity (ρ) are reciprocals of each other: κ=1/ρ. Because a superconductor's resistivity ρ→0, its conductivity κ→∞ (infinite) — current can flow through it indefinitely without any energy loss as heat.
✓Final answerInfinite (conductivity → ∞, since resistivity = 0).
- CBSE 2025Set ANNUAL1 markMCQQ.SI unit of resistivity (specific resistance) is -(a) Ω(b) Ω^-1(c) Ωm(d) Ωm^-1
›Reveal solutionSolution
Resistivity (specific resistance) has SI unit ohm-metre (Ωm).
Resistance of a conductor is related to its resistivity by:
R = rho x (l/A)
where l is length (m) and A is cross-sectional area (m^2). Rearranging:
rho = R x A / l
Units: rho = (ohm) x (m^2) / (m) = ohm x m = Ωm
This is why resistivity is also called 'specific resistance' - it is numerically the resistance of a conductor of unit length and unit cross-sectional area.
✓Final answer(c) Ωm.
- CBSE 2024Set D1 markMCQQ.Which of the following has the highest molar electrical conductance in aqueous solution?(a) [Pt(NH3)6]Cl4(b) [Pt(NH3)5Cl]Cl3(c) [Pt(NH3)4Cl2]Cl2(d) [Pt(NH3)3Cl3]Cl
›Reveal solutionSolution
Molar conductance rises with the number of ions produced on dissociation. [Pt(NH3)6]Cl4 gives 5 ions, the most of the options, so it conducts best.
Count the ions each complex furnishes in water (only the counter-ions outside the coordination sphere ionise):
- [Pt(NH3)6]Cl4 -> [Pt(NH3)6]4+ + 4 Cl- => 5 ions
- [Pt(NH3)5Cl]Cl3 -> [Pt(NH3)5Cl]3+ + 3 Cl- => 4 ions
- [Pt(NH3)4Cl2]Cl2 -> [Pt(NH3)4Cl2]2+ + 2 Cl- => 3 ions
- [Pt(NH3)3Cl3]Cl -> [Pt(NH3)3Cl3]+ + Cl- => 2 ions
More ions and higher ionic charges mean higher molar conductance, so [Pt(NH3)6]Cl4 (5 ions) has the maximum.
✓Final answer(A) [Pt(NH3)6]Cl4.
- CBSE 2024Set D1 markMCQQ.The cell constant of a conductivity cell is(a) l/A(b) A/l(c) l.A(d) R/A
›Reveal solutionSolution
Cell constant = l/A (distance between electrodes ÷ electrode area), unit cm^-1.
Conductance G of a solution in a conductivity cell is G = kappa (A/l), where kappa is conductivity, A is the electrode area and l is the distance between the electrodes.
Rearranging, kappa = G (l/A). The geometric factor (l/A) is called the CELL CONSTANT because it depends only on the fixed geometry of the cell.
Since resistance R = 1/G, we also write kappa = (1/R)(l/A). The cell constant l/A has the unit m^-1 or cm^-1.
✓Final answer(a) l/A — cell constant = distance between electrodes / area of cross-section.
- CBSE 2024Set B1 markMCQQ.The unit of cell constant is(a) ohm cm(b) cm^-1(c) cm(d) ohm^-1 cm^-1
›Reveal solutionSolution
Cell constant G* = l/A (distance between electrodes divided by their area), so its unit works out to cm^-1.
For a conductivity cell, the cell constant is defined as:
G∗=Al
where l is the distance between the two electrodes (in cm) and A is the area of cross-section of the electrodes (in cm^2).
Units: cm / cm^2 = cm^-1.
The cell constant is used to convert the measured conductance of a solution into its conductivity: kappa = G* x (measured conductance).
✓Final answer(b) cm^-1.
- CBSE 2024Set ANNUAL1 markMCQQ.SI unit of conductivity is -(a) S(b) Ω(c) S cm^-1(d) S m^-1
›Reveal solutionSolution
Conductivity kappa = conductance x (l/A); in SI units this comes out as siemens per metre (S m^-1).
Conductance G has the unit siemens (S), where S = ohm^-1 = A/V.
Conductivity kappa is defined as kappa = G x (l/A), where l is the length between electrodes (in metres) and A is the cross-sectional area (in square metres).
So the unit of kappa = S x (m/m^2) = S m^-1.
Option (a) S is just the unit of conductance, not conductivity.
Option (b) ohm is the unit of resistance.
Option (c) S cm^-1 is a practically convenient (CGS-based) unit but not the SI unit, since the SI base unit of length is the metre, not the centimetre.
✓Final answer(d) S m^-1.
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