Concept: Molar Conductivity

Molar conductivity $\Lambda_m$ is defined as $\Lambda_m = \frac{\kappa}{c}$, where $\kappa$ is the conductivity in S m$^{-1}$ and $c$ is the concentration in mol m$^{-3}$. For strong electrolytes like NaCl, $\Lambda_m$ varies linearly with $\sqrt{c}$ at low concentrations, and the intercept at $c = 0$ gives the limiting molar conductivity $\Lambda_m^0$.

Step 1: Convert units

Concentration in M (mol L$^{-1}$) must be converted to mol m$^{-3}$:

$1 \text{ M} = 1000 \text{ mol m}^{-3}$.

So $c$ values: 1, 10, 20, 50, 100 mol m$^{-3}$.

Step 2: Calculate $\Lambda_m$ for each concentration

Using $\Lambda_m = \frac{\kappa}{c}$ (with $\kappa$ in S m$^{-1}$ and $c$ in mol m$^{-3}$), and noting $\kappa$ is given as $10^2 \times \kappa$, so actual $\kappa = (\text{table value}) \times 10^{-2}$ S m$^{-1}$.

Step 3: Plot $\Lambda_m$ vs $\sqrt{c}$

$\sqrt{c}$ values: 1, 3.162, 4.472, 7.071, 10 (in $\sqrt{\text{mol m}^{-3}}$). …

The key idea here is that molar conductivity $\Lambda_m$ measures how well a solution conducts electricity per mole of electrolyte. As concentration decreases, ions move more freely because interionic attractions weaken. By plotting $\Lambda_m$ against $\sqrt{c}$ and extrapolating to zero concentration, we find $\Lambda_m^0$ — the conductivity at infinite dilution where ions are completely independent.

For strong electrolytes like NaCl, the Debye-Hückel-Onsager theory predicts a linear relationship between $\Lambda_m$ and $\sqrt{c}$ at low concentrations. This linearity lets us extrapolate reliably.

1. Convert units and calculate $\Lambda_m$ for each concentration

Molar conductivity is defined as:

where $\kappa$ is in $\text{S m}^{-1}$ and $c$ is in $\text{mol m}^{-3}$. But the table gives $\kappa$ as $10^2 \times \kappa$ in $\text{S m}^{-1}$, so actual $\kappa = (\text{table value}) \times 10^{-2}\ \text{S m}^{-1}$.

Also, concentration is given in M (mol/L), which is $\text{mol dm}^{-3}$. To convert to $\text{mol m}^{-3}$, multiply by $1000$:

Let's compute for each row:

For $c = 0.001\ \text{M}$:

• $c = 0.001 \times 1000 = 1.0\ \text{mol m}^{-3}$
• $\kappa = 1.237 \times 10^{-2} = 0.01237\ \text{S m}^{-1}$
• $\Lambda_m = \frac{0.01237}{1.0} = 0.01237\ \text{S m}^2\ \text{mol}^{-1}$

But molar conductivity is usually expressed in $\text{S cm}^2\ \text{mol}^{-1}$. Since $1\ \text{S m}^2 = 10^4\ \text{S cm}^2$:

Similarly for $c = 0.010\ \text{M}$:

• $c = 10\ \text{mol m}^{-3}$
• $\kappa = 11.85 \times 10^{-2} = 0.1185\ \text{S m}^{-1}$
• $\Lambda_m = \frac{0.1185}{10} = 0.01185\ \text{S m}^2\ \text{mol}^{-1} = 118.5\ \text{S cm}^2\ \text{mol}^{-1}$

For $c = 0.020\ \text{M}$:

• $c = 20\ \text{mol m}^{-3}$
• $\kappa = 23.15 \times 10^{-2} = 0.2315\ \text{S m}^{-1}$
• $\Lambda_m = \frac{0.2315}{20} = 0.011575\ \text{S m}^2\ \text{mol}^{-1} = 115.75\ \text{S cm}^2\ \text{mol}^{-1}$

For $c = 0.050\ \text{M}$:

• $c = 50\ \text{mol m}^{-3}$
• $\kappa = 55.53 \times 10^{-2} = 0.5553\ \text{S m}^{-1}$
• $\Lambda_m = \frac{0.5553}{50} = 0.011106\ \text{S m}^2\ \text{mol}^{-1} = 111.06\ \text{S cm}^2\ \text{mol}^{-1}$

For $c = 0.100\ \text{M}$:

• $c = 100\ \text{mol m}^{-3}$
• $\kappa = 106.74 \times 10^{-2} = 1.0674\ \text{S m}^{-1}$
• $\Lambda_m = \frac{1.0674}{100} = 0.010674\ \text{S m}^2\ \text{mol}^{-1} = 106.74\ \text{S cm}^2\ \text{mol}^{-1}$

2. Tabulate $\Lambda_m$ and $\sqrt{c}$

3. Plot and extrapolate …

Method: Kohlrausch’s Law & Extrapolation Method

This method uses the relationship between molar conductivity ($\Lambda_m$) and concentration ($c$) for strong electrolytes, followed by graphical extrapolation to infinite dilution.

Step 1: Calculate $\Lambda_m$ for each concentration

Formula:

Where:

• $\kappa$ = conductivity (in $S\ m^{-1}$)
• $c$ = concentration (in $mol\ m^{-3}$)

Important: Convert concentration from M ($mol\ L^{-1}$) to $mol\ m^{-3}$:

Step 2: Perform the calculations

For 0.001 M:

• $c = 0.001 \times 1000 = 1\ mol\ m^{-3}$
• $\kappa = 1.237 \times 10^{-2}\ S\ m^{-1}$
• $\Lambda_m = \frac{1.237 \times 10^{-2}}{1} = 1.237 \times 10^{-2}\ S\ m^2\ mol^{-1}$

Similarly for all concentrations:

Step 3: Calculate $\sqrt{c}$ values

Step 4: Plot $\Lambda_m$ vs $\sqrt{c}$

• X-axis: $\sqrt{c}$ (in $M^{1/2}$)
• Y-axis: $\Lambda_m$ (in $S\ m^2\ mol^{-1}$) …

Common Mistakes & How to Avoid Them

Mistake 1: Mishandling the "$10^2 \times \kappa$" Table Header

The error: Students read the table value (e.g., $1.237$) as $\kappa$ itself, instead of realising the header says $10^2 \times \kappa$, so the actual conductivity is the table value $\times 10^{-2}$.

Why it happens: Tables in NCERT-style problems often report a scaled quantity to keep the numbers tidy, and it's easy to skip past the header notation under time pressure.

How to avoid:

• Always read the column header literally: if it says $10^2 \times \kappa\ /\ \text{S m}^{-1}$, then $\kappa = (\text{table value}) \times 10^{-2}\ \text{S m}^{-1}$.
• For $c = 0.001\ M$: table value is $1.237$, so $\kappa = 1.237 \times 10^{-2}\ \text{S m}^{-1}$, not $1.237\ \text{S m}^{-1}$.

Mistake 2: Forgetting to Convert Concentration from mol L⁻¹ to mol m⁻³

The error: Dividing $\kappa$ (in S m⁻¹) directly by $c$ in mol L⁻¹ (i.e., $0.001$, $0.01$, etc.) without converting to SI concentration units, which gives an answer 1000× too large.

How to avoid:

• Since $1\ \text{M} = 1000\ \text{mol m}^{-3}$, always convert first: $c\ (\text{mol m}^{-3}) = c\ (\text{M}) \times 1000$.
• For $c = 0.001\ M$: $c = 1\ \text{mol m}^{-3}$.

Mistake 3: Mixing S m² mol⁻¹ and S cm² mol⁻¹ Without Converting

The error: Computing $\Lambda_m$ in SI units ($\text{S m}^2\ \text{mol}^{-1}$, which comes out as a small number like $0.0124$) and then comparing it directly against a $\Lambda_m^0$ value quoted in $\text{S cm}^2\ \text{mol}^{-1}$ (typically in the hundreds) without converting, making the numbers look wildly inconsistent.

How to avoid:

• Remember: $1\ \text{S m}^2\ \text{mol}^{-1} = 10^4\ \text{S cm}^2\ \text{mol}^{-1}$.
• Pick ONE unit system and stick with it for every row of the table and for the extrapolated intercept.

Mistake 4: Plotting $\Lambda_m$ Against $c$ Instead of $\sqrt{c}$

The error: For a strong electrolyte like NaCl, students plot $\Lambda_m$ directly against concentration $c$ and try to extrapolate — but Kohlrausch's law says the linear relationship is with $\sqrt{c}$, not $c$ itself, so extrapolating the wrong plot gives a wrong intercept.

How to avoid:

• Always compute $\sqrt{c}$ for each row first.
• Plot $\Lambda_m$ (y-axis) against $\sqrt{c}$ (x-axis) — only this plot is a straight line for a strong electrolyte, per $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$.

Mistake 5: Estimating the Intercept from Only Two (Often the Closest) Data Points …