From resistance to conductance
Conduction of electricity through an electrolytic solution is best understood by first recalling the electrical properties of any conductor. The electrical resistance, $R$, is measured in ohm ($\Omega$), which in SI base units equals $(\mathrm{kg\,m^2})/(\mathrm{s^3\,A^2})$. As you learnt in physics, $R$ is directly proportional to the length $l$ of the conductor and inversely proportional to its area of cross-section $A$:
$$R = \rho\,\frac{l}{A}$$
The proportionality constant $\rho$ (rho) is the resistivity (specific resistance). Its SI unit is ohm metre ($\Omega\,\mathrm{m}$); the submultiple $\Omega\,\mathrm{cm}$ is also common, with $1\,\Omega\,\mathrm{m} = 100\,\Omega\,\mathrm{cm}$. Physically, resistivity is the resistance of a sample one metre long with a cross-section of one square metre.
The reciprocal of resistance is conductance, $G$, and the reciprocal of resistivity is conductivity, $\kappa$ (kappa):
$$G = \frac{1}{R} = \frac{\kappa A}{l}, \qquad \kappa = \frac{1}{\rho}$$
The SI unit of conductance is the siemens (S), equal to $\Omega^{-1}$ (the older name is mho). IUPAC recommends the terms resistivity and conductivity over the older "specific" terms; we follow that convention throughout.
| Quantity | Symbol | Defining relation | SI unit | Common unit |
|---|---|---|---|---|
| Resistance | $R$ | $\rho\, l/A$ | Ω | Ω |
| Resistivity | $\rho$ | $RA/l$ | Ω m | Ω cm |
| Conductance | $G$ | $1/R$ | S | S (= Ω⁻¹) |
| Conductivity | $\kappa$ | $1/\rho = G\,(l/A)$ | S m⁻¹ | S cm⁻¹ |
Conductivity and the cell constant
Pure water itself conducts feebly because it contains only $\sim 10^{-7}\,\mathrm{M}$ of $\ce{H+}$ and $\ce{OH-}$ ions, giving a conductivity of about $3.5 \times 10^{-5}\,\mathrm{S\,m^{-1}}$. When an electrolyte such as $\ce{KCl}$ dissolves, $\ce{KCl -> K+ + Cl-}$, it furnishes its own ions and the conductivity rises sharply. Conduction by ions in solution is called electrolytic (ionic) conductance, and the conductivity of an ionic solution depends on the nature of the electrolyte, the size and solvation of the ions, the nature and viscosity of the solvent, the concentration of the electrolyte, and the temperature (κ rises with temperature).
For a column of solution of length $l$ confined between two electrodes of area $A$, the resistance is again $R = \rho\,l/A$. The geometric factor $l/A$ is the cell constant, denoted $G^{*}$:
$$G^{*} = \frac{l}{A} = R\,\kappa$$
The cell constant has the dimension of length$^{-1}$ (units $\mathrm{cm^{-1}}$ or $\mathrm{m^{-1}}$). Because measuring $l$ and $A$ directly is both inconvenient and unreliable, $G^{*}$ is determined by filling the cell with a solution of accurately known conductivity — almost always $\ce{KCl}$, whose conductivity is tabulated at several concentrations and temperatures — and reading the resistance. Once $G^{*}$ is fixed for a given cell, the conductivity of any solution measured in it follows from:
$$\kappa = \frac{G^{*}}{R} = \text{cell constant} \times \text{conductance}$$
Cell constant has units of cm⁻¹, not cm
Students routinely write the cell constant with units of length. Since $G^{*} = l/A$ has dimensions of (length)/(length)², its unit is the reciprocal of length. A 2023 NEET item used exactly this: $G^{*} = \kappa \times R = 0.0210 \times 60 = 1.26\,\mathrm{cm^{-1}}$.
Remember: $G^{*} = \kappa R$ (conductivity × resistance), so its unit is S cm⁻¹ × Ω = cm⁻¹.
Molar conductivity (Λm)
Conductivity alone does not allow a fair comparison between electrolytes, because solutions at the same concentration dissociate into different numbers and types of ions. A more physically meaningful quantity is the molar conductivity, $\Lambda_m$ (lambda), which scales the conductivity to one mole of electrolyte:
$$\Lambda_m = \frac{\kappa}{c}$$
Here $c$ is the molar concentration. Geometrically, $\Lambda_m$ is the conductance of the whole volume $V$ of solution that contains exactly one mole of electrolyte, held between electrodes that are a unit distance apart but large enough in area to enclose that entire volume. Because $l = 1$ and $A = V$ in those units:
$$\Lambda_m = \kappa V$$
If $\kappa$ is in $\mathrm{S\,m^{-1}}$ and $c$ in $\mathrm{mol\,m^{-3}}$, then $\Lambda_m$ comes out in $\mathrm{S\,m^2\,mol^{-1}}$. In the more familiar centimetre system, with $\kappa$ in $\mathrm{S\,cm^{-1}}$ and molarity in $\mathrm{mol\,L^{-1}}$, the working formula is:
$$\Lambda_m\,(\mathrm{S\,cm^2\,mol^{-1}}) = \frac{\kappa\,(\mathrm{S\,cm^{-1}}) \times 1000\,(\mathrm{cm^3\,L^{-1}})}{\text{molarity}\,(\mathrm{mol\,L^{-1}})}$$
Measuring conductivity: AC and the Wheatstone bridge
An unknown resistance is normally measured on a Wheatstone bridge, but an ionic solution presents two difficulties. First, passing direct current for a prolonged period electrolyses the solution and changes its composition near the electrodes. Second, a liquid cannot be wired into the bridge like a metal conductor. Both problems are solved by (i) using an alternating current source — an oscillator in the audio range, 550 to 5000 cycles per second, which reverses direction so quickly that no net electrolysis occurs — and (ii) confining the solution in a purpose-built conductivity cell with two platinised platinum electrodes.
In the bridge, two fixed resistances $R_3$ and $R_4$ and a variable resistance $R_1$ are balanced against the cell of unknown resistance $R_2$. A detector $P$ (a headphone or electronic null detector) shows no current when the bridge is balanced, and then:
$$R_2 = \frac{R_1\,R_4}{R_3}$$
Modern conductivity meters read $G$ or $R$ directly; combined with the known cell constant they give $\kappa = G^{*}/R$ at once.
Once you can find $\Lambda^{\circ}_m$ for strong electrolytes, the next step is splitting it into ionic contributions. See Kohlrausch's law of independent migration of ions.
Units and conversions you must memorise
More NEET marks are lost to unit slips here than to conceptual errors. The two systems — SI (metre) and the practical (centimetre) — coexist in textbooks and questions, so you must convert fluently between them.
| Quantity | SI unit | Practical unit | Conversion |
|---|---|---|---|
| Conductivity ($\kappa$) | S m⁻¹ | S cm⁻¹ | $1\,\mathrm{S\,cm^{-1}} = 100\,\mathrm{S\,m^{-1}}$ |
| Molar conductivity ($\Lambda_m$) | S m² mol⁻¹ | S cm² mol⁻¹ | $1\,\mathrm{S\,m^2\,mol^{-1}} = 10^4\,\mathrm{S\,cm^2\,mol^{-1}}$ |
| Cell constant ($G^{*}$) | m⁻¹ | cm⁻¹ | $1\,\mathrm{cm^{-1}} = 100\,\mathrm{m^{-1}}$ |
| Concentration ($c$) | mol m⁻³ | mol L⁻¹ | $1\,\mathrm{mol\,L^{-1}} = 1000\,\mathrm{mol\,m^{-3}}$ |
Forgetting the factor of 1000 in Λm = κ/c
In the centimetre system $\kappa$ is in $\mathrm{S\,cm^{-1}}$ but molarity is in $\mathrm{mol\,L^{-1}}$, and $1\,\mathrm{L} = 1000\,\mathrm{cm^3}$. You therefore multiply by 1000: $\Lambda_m = \kappa \times 1000 / \text{molarity}$. Omitting it makes $\Lambda_m$ a thousand times too small.
Pure SI ($\kappa$ in S m⁻¹, $c$ in mol m⁻³) needs no extra factor — only the mixed cm/L system does.
Variation of κ and Λm with concentration
Both $\kappa$ and $\Lambda_m$ change with concentration, but in opposite directions on dilution — the single most examined contrast in this subtopic.
Conductivity always decreases on dilution, for both strong and weak electrolytes. The reason is direct: $\kappa$ is the conductance of one unit volume of solution, and on dilution the number of current-carrying ions per unit volume falls, so less current is carried per unit volume.
Molar conductivity always increases on dilution. Although $\kappa$ drops, $\Lambda_m = \kappa V$ refers to all the ions from one mole of electrolyte, spread over a volume $V$ that grows on dilution. The increase in $V$ more than compensates for the fall in $\kappa$, so $\Lambda_m$ rises. As $c \to 0$, $\Lambda_m$ approaches a limiting value called the limiting molar conductivity, $\Lambda^{\circ}_m$.
| On dilution (c decreases) | Conductivity κ | Molar conductivity Λm |
|---|---|---|
| Direction of change | Decreases | Increases |
| Reason | Fewer ions per unit volume | Volume $V$ per mole grows faster than κ falls |
| Limiting value | → that of pure solvent | → $\Lambda^{\circ}_m$ (limiting molar conductivity) |
| True for | Strong & weak | Strong & weak |
Strong versus weak electrolytes
The manner in which $\Lambda_m$ rises on dilution distinguishes the two classes sharply, and the standard plot is $\Lambda_m$ against $\sqrt{c}$.
Strong electrolytes (such as $\ce{KCl}$, $\ce{NaCl}$, $\ce{CaCl2}$) are essentially fully dissociated at all concentrations. Their $\Lambda_m$ increases only slowly with dilution because the few remaining inter-ionic interactions weaken. The dependence is linear in $\sqrt{c}$ (Debye–Hückel–Onsager behaviour):
$$\Lambda_m = \Lambda^{\circ}_m - A\sqrt{c}$$
A plot of $\Lambda_m$ versus $\sqrt{c}$ is a straight line whose intercept at $\sqrt{c}=0$ gives $\Lambda^{\circ}_m$ directly and whose slope is $-A$. The constant $A$ depends only on the electrolyte type (1-1, 2-1, 2-2, etc.) for a given solvent and temperature, so $\ce{NaCl}$, $\ce{CaCl2}$ and $\ce{MgSO4}$ illustrate the three types with their own $A$ values.
Weak electrolytes (such as $\ce{CH3COOH}$) are only partly dissociated, $\ce{CH3COOH <=> CH3COO^- + H+}$, and their degree of dissociation $\alpha$ rises steeply only near infinite dilution. Consequently $\Lambda_m$ is small at moderate concentrations and shoots up sharply as $c \to 0$. The curve does not follow a clean straight line, so $\Lambda^{\circ}_m$ for a weak electrolyte cannot be obtained by extrapolation — it must be computed indirectly through Kohlrausch's law.
| Feature | Strong electrolyte (KCl) | Weak electrolyte (CH₃COOH) |
|---|---|---|
| Dissociation | Almost complete at all c | Partial; α rises near infinite dilution |
| Λm vs √c shape | Near-straight line | Gentle, then steep rise near c → 0 |
| Equation | $\Lambda_m = \Lambda^{\circ}_m - A\sqrt{c}$ | No simple linear law |
| How Λ°m is found | Extrapolate the straight line | Kohlrausch's law (indirect) |
| Cause of Λm rise | Weaker inter-ionic forces | Increasing degree of dissociation |
Worked numericals
The resistance of a cell filled with $0.1\,\mathrm{mol\,L^{-1}}$ KCl is $100\,\Omega$ (its conductivity is $1.29\,\mathrm{S\,m^{-1}}$). The same cell filled with $0.02\,\mathrm{mol\,L^{-1}}$ KCl reads $520\,\Omega$. Find the conductivity and molar conductivity of the dilute solution.
Cell constant $G^{*} = \kappa \times R = 1.29\,\mathrm{S\,m^{-1}} \times 100\,\Omega = 129\,\mathrm{m^{-1}}$.
Conductivity of the dilute solution: $\kappa = G^{*}/R = 129/520 = 0.248\,\mathrm{S\,m^{-1}}$.
With $c = 0.02\,\mathrm{mol\,L^{-1}} = 20\,\mathrm{mol\,m^{-3}}$, $\;\Lambda_m = \kappa/c = (0.248)/(20) = 124 \times 10^{-4}\,\mathrm{S\,m^2\,mol^{-1}}$, i.e. $124\,\mathrm{S\,cm^2\,mol^{-1}}$.
A column of $0.05\,\mathrm{mol\,L^{-1}}$ NaOH of diameter $1\,\mathrm{cm}$ and length $50\,\mathrm{cm}$ has resistance $5.55 \times 10^{3}\,\Omega$. Find its conductivity and molar conductivity.
Area $A = \pi r^2 = 3.14 \times (0.5)^2 = 0.785\,\mathrm{cm^2}$; length $l = 50\,\mathrm{cm}$.
Resistivity $\rho = RA/l = (5.55\times10^3 \times 0.785)/50 = 87.135\,\Omega\,\mathrm{cm}$, so $\kappa = 1/\rho = 0.01148\,\mathrm{S\,cm^{-1}}$.
$\Lambda_m = \kappa \times 1000/\text{molarity} = (0.01148 \times 1000)/0.05 = 229.6\,\mathrm{S\,cm^2\,mol^{-1}}$.
Conductance of electrolytic solutions in one screen
- $G = 1/R$ (siemens); $\kappa = 1/\rho = G\,(l/A)$; cell constant $G^{*} = l/A = \kappa R$, unit cm⁻¹.
- Molar conductivity $\Lambda_m = \kappa/c = \kappa V$; in cm/L units multiply by 1000: $\Lambda_m = \kappa \times 1000/\text{molarity}$.
- Resistance is measured with AC (to avoid electrolysis) on a Wheatstone bridge: $R_2 = R_1 R_4/R_3$. Cell constant is calibrated with KCl.
- On dilution: $\kappa$ decreases (fewer ions per unit volume); $\Lambda_m$ increases (volume per mole grows), tending to $\Lambda^{\circ}_m$.
- Strong electrolyte: $\Lambda_m = \Lambda^{\circ}_m - A\sqrt{c}$, straight line → $\Lambda^{\circ}_m$ by extrapolation. Weak electrolyte: steep rise near $c\to0$, $\Lambda^{\circ}_m$ via Kohlrausch's law.
- Units: $1\,\mathrm{S\,cm^{-1}} = 100\,\mathrm{S\,m^{-1}}$; $1\,\mathrm{S\,m^2\,mol^{-1}} = 10^4\,\mathrm{S\,cm^2\,mol^{-1}}$.