When 0.1 mol $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$, 0.2 mol of $AgCl$ are obtained. The conductivity of solution will correspond to (i) 1:3 electrolyte (ii) 1:2 electrolyte (iii) 1:1 electrolyte (iv) 3:1 electrolyte

When 0.1 mol CoCl₃(NH₃)₅ is treated with excess of AgNO₃, 0.2 mol of AgCl are obtained. The conductivity of so

NCERT Exemplar · Q3

CoCl_3(NH_3)_5

(i) 1:3 electrolyte

(ii) 1:2 electrolyte

(iii) 1:1 electrolyte

(iv) 3:1 electrolyte

CBSEMCQ· 1mImportance★★★★★

50% · 51/101 Questions

✓ Free question

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.

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).
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

Important

$G = \frac{1}{R}$ and $\sigma = \frac{1}{\rho}$ .
$G = \sigma \frac{A}{L}$ 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 ( $\sigma$ ).

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
$\mu$ = 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 ( $\mu$ large) → electrons move faster for the same push → higher conductivity.

This explains why metals (high $n$ ) are good conductors, and why heating reduces $\sigma$ (more collisions → lower $\mu$ ).

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	$R = \rho L/A$	Geometry + material
4	Conductivity	$\sigma = 1/\rho$	Material's intrinsic ability
5	Key result	$\boxed{G = \sigma \frac{A}{L}}$	Combines material + geometry

Final takeaway:

Conductance $G$ is not just a number — it's the product of how good the material is ( $\sigma$ ) 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.

The complex is $[Co(NH_3)_5Cl]Cl_2$ , which dissociates into 3 ions (1 Co complex cation + 2 Cl⁻ anions), so it behaves as a 1:2 electrolyte. The correct option is (ii).

Why Conductivity and Precipitation Tell the Same Story

When you dissolve a coordination compound in water, its conductivity depends on how many ions it releases. A 1:1 electrolyte (like NaCl) gives 2 ions, a 1:2 electrolyte (like CaCl₂) gives 3 ions, and so on. The precipitation experiment with AgNO₃ is a clever way to count how many chloride ions are free — because only free Cl⁻ ions react with Ag⁺ to form AgCl. Chloride ions that are tightly bound inside the coordination sphere (as ligands) do not precipitate.

So the two experiments — conductivity and precipitation — are two sides of the same coin. If you know how many Cl⁻ ions are free, you know the ionic formula of the complex, and from that you can predict how many total ions it will produce in solution.

Step-by-Step Reasoning

Interpret the precipitation data.

The reaction is:

$\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \downarrow$

From 0.1 mol of the complex, we get 0.2 mol of AgCl. That means 0.2 mol of free Cl⁻ ions were present in the solution.

So per mole of complex, the number of free Cl⁻ ions is:

$\frac{0.2}{0.1} = 2$

Therefore, the complex must have two chloride ions outside the coordination sphere (as counter-ions) and the rest inside as ligands.
Deduce the formula of the complex.

The molecular formula is $CoCl_3(NH_3)_5$ . We have 3 Cl atoms total. Two are free (ionic), so the remaining one must be coordinated to cobalt inside the coordination sphere.

Hence the correct structural formula is:

$[Co(NH_3)_5Cl]Cl_2$

The square brackets enclose the coordination sphere: one Co³⁺ ion, five NH₃ molecules, and one Cl⁻ ligand. Outside the sphere, two Cl⁻ ions balance the +3 charge on the cobalt (since Co³⁺ + 5 neutral NH₃ + 1 Cl⁻ ligand gives a net +2 charge on the complex ion, which is neutralized by the two free Cl⁻ ions).
Determine the number of ions in solution.

When $[Co(NH_3)_5Cl]Cl_2$ dissolves, it dissociates completely into:

$[Co(NH_3)_5Cl]^{2+} + 2Cl^-$

That is 3 ions per formula unit: one complex cation with charge +2, and two chloride anions.
Classify the electrolyte type.

An electrolyte that gives one cation of charge +2 and two anions of charge -1 is called a 1:2 electrolyte (the ratio of cation charge to anion charge is 1:2, but more commonly it refers to the number of ions: 1 cation : 2 anions).

Option (ii) matches this.

Watch out

A common mistake is to think that all 3 chlorides are free because the formula shows 3 Cl atoms. Remember: only chlorides outside the coordination sphere precipitate with Ag⁺. The coordinated Cl⁻ is "invisible" to AgNO₃.

Tip

You can also work backwards from the conductivity: a 1:2 electrolyte gives a molar conductivity roughly double that of a 1:1 electrolyte (for similar ions), but here the precipitation data alone is sufficient — no need for actual conductivity values.

✓Final answer

The correct option is (ii) 1:2 electrolyte .

Common Mistakes & How to Avoid Them

Mistake 1: Counting All Chlorines as Precipitable

The error: Students see the formula $CoCl_3(NH_3)_5$ and assume all 3 chlorine atoms will precipitate with $AgNO_3$ , expecting 0.3 mol of $AgCl$ .

Why it's wrong: Only chloride ions outside the coordination sphere (free ions) react with $AgNO_3$ . Chlorine atoms inside the coordination sphere (bonded to the metal) do not precipitate.

How to avoid: Always distinguish between:

Ionizable chlorines (outside coordination sphere) → precipitate with $Ag^+$
Coordinated chlorines (inside coordination sphere) → do not precipitate

Mistake 2: Misinterpreting the Given Data

The error: Not connecting "0.2 mol $AgCl$ obtained" to the number of free chloride ions.

Why it's wrong: The reaction is:

CoCl_3(NH_3)_5 + AgNO_3 \rightarrow AgCl \downarrow + \text{other products}

0.2 mol $AgCl$ means only 2 out of 3 chlorines are free $Cl^-$ ions.

How to avoid: Write the precipitation equation:

Cl^- + Ag^+ \rightarrow AgCl

So moles of $AgCl$ = moles of free $Cl^-$ = 2 (from 0.2 mol obtained from 0.1 mol complex).

Mistake 3: Confusing Conductivity with Number of Ions

The error: Thinking conductivity depends only on the number of ions, ignoring charge.

Why it's wrong: Conductivity depends on both number and charge of ions. A 1:2 electrolyte (like $CaCl_2$ ) gives 3 ions total, but the conductivity pattern differs from a 1:3 electrolyte.

How to avoid: Remember:

1:1 electrolyte → 2 ions (e.g., $NaCl$ )
1:2 electrolyte → 3 ions (e.g., $MgCl_2$ )
1:3 electrolyte → 4 ions (e.g., $AlCl_3$ )

Mistake 4: Forgetting the Complex Cation's Charge

The error: Determining the complex formula but not calculating the charge on the complex ion.

Why it's wrong: The complex is $[Co(NH_3)_5Cl]Cl_2$ (since 2 $Cl^-$ are free). You must find the charge on $[Co(NH_3)_5Cl]^{n+}$ .

How to avoid: Use charge balance:

$Co$ is in +3 oxidation state (common for cobalt complexes)
$NH_3$ is neutral
$Cl$ inside sphere is $-1$
So charge on complex = $+3 + (-1) = +2$

Thus the complex is $[Co(NH_3)_5Cl]^{2+}$ with 2 $Cl^-$ ions → 1:2 electrolyte.

Mistake 5: Picking 1:3 Electrolyte Out of Habit

The error: Seeing $CoCl_3$ and automatically choosing 1:3 without checking coordination.

Why it's wrong: Coordination chemistry changes the number of free ions. The formula $CoCl_3(NH_3)_5$ is not the same as $CoCl_3$ in solution.

How to avoid: Always check:

How many $Cl^-$ precipitate with $AgNO_3$ ?
That number = number of free $Cl^-$ ions
The rest are inside the coordination sphere

Quick Summary Checklist

Step	Action
1	Moles of $AgCl$ = moles of free $Cl^-$
2	Subtract free $Cl^-$ from total $Cl$ to get coordinated $Cl$
3	Write correct complex formula: $[Co(NH_3)_5Cl]Cl_2$
4	Find charge on complex ion using oxidation states
5	Count total ions and their charges → determine electrolyte type

Final answer: The complex gives $[Co(NH_3)_5Cl]^{2+}$ and $2Cl^-$ → 1:2 electrolyte → Option (ii).

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 27 on this concept.

CBSE 2026Set A1 markMCQ
$Q. On increasing dilution, the specific conductance of an electrolyte$
(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 markMCQ
$Q. The number of ions in aqueous solution of [Co(NH3)5Cl]Cl2 is$
(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 markMCQ
$Q. The unit of cell constant is:$
(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): $G^{*} = \dfrac{l}{A}$ . Since l has units of cm and A has units of cm^2, the cell constant has units of $\dfrac{cm}{cm^2} = cm^{-1}$ .

This is distinct from conductivity κ (specific conductance), whose unit is $\Omega^{-1}cm^{-1}$ (obtained as $G^{*}\times$ the measured conductance $1/R$ , unit $\Omega^{-1}$ ), and from molar conductivity Λm, whose unit is $\Omega^{-1}cm^{2}mol^{-1}$ . 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 markMCQ
$Q. The unit of specific conductivity is:$
(a) ohm⁻¹
(b) ohm⁻¹ cm⁻¹
(c) ohm cm
(d) ohm cm⁻¹
›Reveal solutionSolution

Specific conductance ( $\kappa$ ) is measured in ohm⁻¹ cm⁻¹ (S cm⁻¹).

Specific conductivity ( $\kappa$ ), 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 $\kappa = G \times (l/A)$ , 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 markMCQ
$Q. Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : For measuring resistance of an ionic solution an AC source is used. Reason (R) : Concentration of ionic solution will change if DC source is used.$
›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 $\text{NaCl}$ solution, $\text{Cl}^-$ ions get oxidized at the anode ( $2\text{Cl}^- \to \text{Cl}_2 + 2e^-$ ) and $\text{H}^+$ from water gets reduced at the cathode ( $2\text{H}^+ + 2e^- \to \text{H}_2$ ). 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 out

Even 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 answer
$The correct option is (A) : Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.$
CBSE 2025Set D1 markMCQ
$Q. The unit of specific conductance is$
(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 mark
$Q. Write the value of conductivity of superconductor.$
›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 ( $\kappa$ ) and resistivity ( $\rho$ ) are reciprocals of each other: $\kappa = 1/\rho$ . Because a superconductor's resistivity $\rho \to 0$ , its conductivity $\kappa \to \infty$ (infinite) — current can flow through it indefinitely without any energy loss as heat.

✓Final answer
$Infinite (conductivity \to \infty, since resistivity = 0).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. SI unit of resistivity (specific resistance) is -$
(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 markMCQ
$Q. Which of the following has the highest molar electrical conductance in aqueous solution?$
(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 markMCQ
$Q. The cell constant of a conductivity cell is$
(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 markMCQ
$Q. The unit of cell constant is$
(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:

$G^{*} = \frac{l}{A}$

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 markMCQ
$Q. SI unit of conductivity is -$
(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
$(d) S m^-1.$

←Previous questionQ2 — The colour of the coordination compounds depends on the crystal field …→Next questionQ4 — When 1 mol CrCl₃ · 6H₂O is treated with excess of AgNO₃, 3 mol of AgCl…

Q.When 0.1 mol $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$ , 0.2 mol of $AgCl$ are obtained. The conductivity of solution will correspond to

Concept understanding — Conductance And Conductivity

From Resistance to Conductance: Flipping the Idea

Ohm's Law in Conductance Form

Now, Conductivity: The Material's Intrinsic Property

The Big Picture in One Table

Intuitive Analogy

Key Takeaway for Exams

Why this formula?

Conductance and Conductivity: Why the Formulas Hold

1. The Core Idea: How Easily Does Current Flow?

2. Ohm's Law in Terms of Conductance

3. From Resistance to Conductivity: The Geometry Factor

4. The Key Formula: Conductance in Terms of Conductivity

5. Microscopic Derivation (Why $\sigma$ Exists)

6. Summary: The Logical Chain

Why Conductivity and Precipitation Tell the Same Story

Step-by-Step Reasoning

Method: Conductance & Coordination Compound Analysis

Step-by-step reasoning

Final Answer

Common Mistakes & How to Avoid Them

Mistake 1: Counting All Chlorines as Precipitable

Mistake 2: Misinterpreting the Given Data

Mistake 3: Confusing Conductivity with Number of Ions

Mistake 4: Forgetting the Complex Cation's Charge

Mistake 5: Picking 1:3 Electrolyte Out of Habit

Quick Summary Checklist

Why DC causes problems

Why AC solves this

Evaluating the statements

Q.When 0.1 mol CoCl3(NH3)5CoCl_3(NH_3)_5CoCl3​(NH3​)5​ is treated with excess of AgNO3AgNO_3AgNO3​, 0.2 mol of AgClAgClAgCl are obtained. The conductivity of solution will correspond to

Why Conductivity and Precipitation Tell the Same Story

Step-by-Step Reasoning

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Q.When 0.1 mol $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$ , 0.2 mol of $AgCl$ are obtained. The conductivity of solution will correspond to

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