Q.The conductivity of $0.20\ M$ solution of KCl at 298 K is $0.0248\ S\ cm^{-1}$. Calculate its molar conductivity.

Concept: Molar Conductivity — the conductivity of all ions produced from one mole of electrolyte in solution.

The formula is:

where $\kappa$ is conductivity in $\text{S cm}^{-1}$, $C$ is molarity in $\text{mol L}^{-1}$, and the factor 1000 converts $\text{cm}^3$ to $\text{L}$.

Step 1: Identify the given values.

$\kappa = 0.0248\ \text{S cm}^{-1}$, $C = 0.20\ \text{mol L}^{-1}$.

Step 2: Substitute into the formula. …

Molar conductivity ($\Lambda_m$) is a way to compare how well different electrolytes conduct electricity, independent of their concentration. Think of it this way: conductivity ($\kappa$) tells you how much current flows through a given volume of solution. But if you have a more concentrated solution, there are simply more ions present to carry charge — so the conductivity goes up just because there are more carriers. Molar conductivity removes this "crowding" effect by normalising to a fixed amount of electrolyte (one mole). It answers: If I had exactly one mole of this electrolyte dissolved, how well would it conduct?

The formula is straightforward:

where $\kappa$ is the measured conductivity (in $\text{S cm}^{-1}$) and $c$ is the molar concentration (in $\text{mol L}^{-1}$). The tricky part — and the reason many students slip up — is the units. Conductivity is usually given in $\text{S cm}^{-1}$, but concentration is in $\text{mol L}^{-1}$. One litre is $1000\ \text{cm}^3$, so you must convert the concentration to $\text{mol cm}^{-3}$ before dividing, or equivalently, multiply by 1000 after dividing. Let's walk through it.

1. Write down what's given.

   $\kappa = 0.0248\ \text{S cm}^{-1}$

   $c = 0.20\ \text{M} = 0.20\ \text{mol L}^{-1}$

2. Convert concentration to $\text{mol cm}^{-3}$.

   Since $1\ \text{L} = 1000\ \text{cm}^3$,

3. Apply the formula.

Dividing:

$\Lambda_m = \frac{0.0248}{2.0 \times 10^{-4}}\ \text{S cm}^2\ \text{mol}^{-1} = 124\ \text{S cm}^2\ \text{mol}^{-1}$ …

Method: Direct Formula Substitution

This is the most straightforward method — simply apply the definition of molar conductivity.

Concept (Why this works)

Molar conductivity ($\Lambda_m$) tells us the conducting power of all ions produced from one mole of an electrolyte. It relates the measured conductivity ($\kappa$) to the concentration ($c$) of the solution.

Formula

Where:

• $\kappa$ = conductivity in S cm⁻¹
• $c$ = concentration in mol L⁻¹
• Factor 1000 converts cm³ to L (since 1 L = 1000 cm³)

Steps

1. Identify given data

   • $\kappa = 0.0248\ \text{S cm}^{-1}$
   • $c = 0.20\ \text{M} = 0.20\ \text{mol L}^{-1}$

2. Apply the formula

3. Simplify

   $\Lambda_m = \frac{24.8}{0.20} = 124$ …

Here are the common mistakes students make when calculating molar conductivity from conductivity data, along with how to avoid each.

1. Forgetting to Convert Units (The Most Common Mistake)

The Mistake:

Students plug the given conductivity ($0.0248\ \text{S cm}^{-1}$) directly into the formula without checking units. The molar conductivity formula requires conductivity in S cm⁻¹ and concentration in mol cm⁻³, but concentration is often given in mol L⁻¹ (or M).

How to Avoid:

Always write the formula first and check every unit.

• Molar conductivity:

where $\kappa$ is conductivity (S cm⁻¹) and $c$ is concentration (mol cm⁻³).

• Given: $c = 0.20\ \text{M} = 0.20\ \text{mol L}^{-1}$ Convert to mol cm⁻³:

• Then:

Key takeaway: Always convert M → mol cm⁻³ by dividing by 1000.

2. Using the Wrong Formula or Confusing Conductivity with Molar Conductivity

The Mistake:

Some students use $\Lambda_m = \kappa \times c$ (multiplying instead of dividing) or confuse $\kappa$ (conductivity) with $\Lambda_m$ (molar conductivity).

How to Avoid:

Remember the definition: Molar conductivity is the conductivity of a solution containing 1 mole of electrolyte placed between electrodes 1 cm apart.

• If conductivity is high for a dilute solution, molar conductivity is large.
• The relationship is inverse: $\Lambda_m = \frac{\kappa}{c}$.

Mnemonic: “Conductivity per mole” → divide by concentration.

3. Ignoring the Temperature Dependence

The Mistake:

Students assume the same formula works at any temperature without noting that conductivity changes with temperature. The problem explicitly states 298 K — using a different temperature’s data or formula would be wrong.

How to Avoid:

Always note the temperature given. For exam problems, use the data as provided. If temperature is not given, assume standard conditions (298 K). Never mix data from different temperatures.

4. Misplacing Decimal Points in Unit Conversion

The Mistake:

When converting $0.20\ \text{M}$ to mol cm⁻³, students sometimes write $0.20/1000 = 0.00020$ but then misplace the decimal in the final division, e.g., $0.0248 / 0.00020 = 1240$ instead of $124$.

How to Avoid:

Use scientific notation for clarity:

• $c = 2.0 \times 10^{-4}\ \text{mol cm}^{-3}$
• $\kappa = 2.48 \times 10^{-2}\ \text{S cm}^{-1}$
• Then: …