Molar and equivalent conductance

Molar conductance. Take one mole of an electrolyte, dissolve it in a solution, and imagine placing that entire solution between two electrodes 1 cm apart. The conductance you'd measure is $\Lambda_m$:

where: $\kappa$ is specific conductance in $\text{S cm}^{-1}$, $C$ is molar concentration in $\text{mol L}^{-1}$, and the factor $1000$ converts litres to $\text{cm}^3$ so $\Lambda_m$ comes out in $\text{S cm}^2\text{mol}^{-1}$. Dividing by $C$ is what "removes" concentration from the picture — $\Lambda_m$ asks how well one mole's worth of ions conducts, regardless of how dilute or concentrated the solution happens to be.

Equivalent conductance. Defined the same way but per gram-equivalent rather than per mole:

where: $C_{eq}$ is normality (equivalents per litre), giving units $\text{S cm}^2\text{eq}^{-1}$. For a 1:1 electrolyte such as NaCl, one mole equals one equivalent, so $\Lambda_m = \Lambda_{eq}$ numerically. For a polyvalent electrolyte such as $\text{CaCl}_2$ or $\text{Al}_2(\text{SO}_4)_3$, one mole supplies more than one equivalent of charge, so $\Lambda_{eq} = \Lambda_m / n$, where $n$ is the total positive (or negative) charge per formula unit — 2 for $\text{CaCl}_2$, 6 for $\text{Al}_2(\text{SO}_4)_3$.

Opposite trends on dilution. As a solution is diluted, $\kappa$ and $\Lambda_m$ move in opposite directions, and this is the single most exam-tested idea in the topic:

• $\kappa$ falls on dilution, simply because fewer ions occupy each cubic centimetre of solution as more solvent is added — less "stuff" per unit volume means less conductance per unit volume.
• $\Lambda_m$ rises on dilution, because it is normalised per mole. As concentration drops, ion–ion attractions weaken (ions move more freely), and for weak electrolytes, dilution also pushes the dissociation equilibrium further toward ions — both effects make each mole of electrolyte conduct better.

Strong vs weak electrolytes on a $\Lambda_m$ vs $\sqrt{C}$ graph. Plotting $\Lambda_m$ against $\sqrt{C}$ (rather than $C$ directly) makes the strong-electrolyte curve come out almost a straight line, which is what makes graphical extrapolation practical:

• Strong electrolytes are already fully dissociated, so $\Lambda_m$ rises only slightly and linearly as $\sqrt{C} \to 0$; the small rise reflects weakening inter-ionic attraction, not new ions appearing. Extrapolating this near-straight line to $C \to 0$ gives $\Lambda_m^\circ$, the limiting molar conductance at infinite dilution, directly by eye.
• Weak electrolytes show a sharp upward curve at low concentration, because dilution shifts the dissociation equilibrium and the degree of dissociation $\alpha$ climbs toward 1, producing many more ions per mole. This curve is too steep and non-linear near the axis to extrapolate reliably, so $\Lambda_m^\circ$ for a weak electrolyte must instead be obtained indirectly, via Kohlrausch's law.
• Degree of dissociation at a given concentration follows from comparing the measured $\Lambda_m$ to the limiting value: $\alpha = \Lambda_m / \Lambda_m^\circ$.

Problem, part (a): The specific conductance of a $0.02\ M$ solution of an electrolyte is $2.48\times10^{-2}\ \text{S cm}^{-1}$. Calculate its molar conductance. Part (b): Separately, a weak monobasic acid HA has $\Lambda_m = 16.2\ \text{S cm}^2\text{mol}^{-1}$ at $0.01\ M$, and its limiting molar conductance (from Kohlrausch's law) is $\Lambda_m^\circ = 390.7\ \text{S cm}^2\text{mol}^{-1}$. Find its degree of dissociation $\alpha$ at this concentration.

Step 1 (part a). Apply the defining ratio: $\Lambda_m = \dfrac{\kappa \times 1000}{C} = \dfrac{2.48\times10^{-2}\ \text{S cm}^{-1} \times 1000\ \text{cm}^3\text{L}^{-1}}{0.02\ \text{mol L}^{-1}}$.

Step 2 (part a). Simplify the numerator first: $2.48\times10^{-2}\times1000 = 24.8\ \text{S cm}^2\text{L}^{-1}$. Then divide by the concentration: $\Lambda_m = \dfrac{24.8}{0.02} = 1240\ \text{S cm}^2\text{mol}^{-1}$.

Step 3 (part b). For the weak acid, apply $\alpha = \Lambda_m/\Lambda_m^\circ = 16.2/390.7 = 0.0415$.

Step 4 (part b). Interpret the result: only about $4.15\%$ of HA is dissociated at $0.01\ M$. This is far below $1$, exactly as expected for a weak electrolyte at an ordinary (non-infinite-dilution) concentration — its $\Lambda_m$ sits well short of $\Lambda_m^\circ$ because most molecules remain undissociated.

Answer: (a) $\Lambda_m = \mathbf{1240\ \text{S cm}^2\text{mol}^{-1}}$ for the $0.02\ M$ solution. (b) $\alpha = \mathbf{0.0415}$ (about $4.15\%$ dissociated) for the weak acid HA at $0.01\ M$.

• Confusing the direction of change on dilution: $\kappa$ falls while $\Lambda_m$ rises. These are opposite trends, and swapping them is the single most common error in this topic.
• Dropping the factor of 1000 when converting from $\kappa$ (effectively per $\text{cm}^3$) to $\Lambda_m$ (per litre-based concentration) — this unit mismatch silently gives an answer 1000 times too small.
• Extrapolating a weak electrolyte's $\Lambda_m$ vs $\sqrt{C}$ graph straight to $C = 0$ as if it were a strong electrolyte's — the steep curvature near the axis makes this invalid; Kohlrausch's law is needed instead.
• Forgetting that $\alpha = \Lambda_m/\Lambda_m^\circ$ can never legitimately exceed 1 — a computed $\alpha > 1$ or a $\Lambda_m$ larger than $\Lambda_m^\circ$ signals mismatched data, not a genuinely more-dissociated electrolyte.

Ask the tutor: if you'd like more practice converting between $\kappa$, $\Lambda_m$ and $\Lambda_{eq}$, or want the Kohlrausch's law derivation of $\Lambda_m^\circ$ walked through step by step.