Question

Consider the following data for KCl solution at a particular temperature:

What is the value of the limiting molar conductivity?

• A. 150.1 S cm\(^2\) mol\(^{-1}\)
• B. 149.2 S cm\(^2\) mol\(^{-1}\)
• C. 151.1 S cm\(^2\) mol\(^{-1}\)
• D. 152.1 S cm\(^2\) mol\(^{-1}\)

Hint:

The Debye-Hückel-Onsager plot is linear with \(\sqrt{C}\) on the x-axis.
Always convert the concentration \(C\) to \(\sqrt{C}\) before doing calculations, as using \(C\) directly is a common mistake in this type of problem.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question provides two data points of concentration and molar conductivity for a strong electrolyte (\(\text{KCl}\)) solution and asks us to calculate its limiting molar conductivity (\(\Lambda_m^\circ\)).

Step 2: Key Formula or Approach:
For a strong electrolyte like \(\text{KCl}\), the variation of molar conductivity (\(\Lambda_m\)) with concentration (\(C\)) is given by the Debye-Hückel-Onsager equation:
\[ \Lambda_m = \Lambda_m^\circ - A \sqrt{C} \]
where:
- \(\Lambda_m^\circ\) is the limiting molar conductivity.
- \(A\) is a constant.
- \(C\) is the concentration.

Step 3: Detailed Explanation:
Let's plug the two data points into the Debye-Hückel-Onsager equation:
1. For the first concentration \(C_1 = 1 \times 10^{-4}\text{ mol L}^{-1}\):
- Molar conductivity \(\Lambda_{m,1} = 149.1\text{ S cm}^2\text{ mol}^{-1}\).
- Calculate \(\sqrt{C_1}\):
\[ \sqrt{C_1} = \sqrt{1 \times 10^{-4}} = 10^{-2} = 0.01 \]
- Substitute into the equation:
\[ 149.1 = \Lambda_m^\circ - A(0.01) \quad \text{--- (Equation 1)} \]
2. For the second concentration \(C_2 = 9 \times 10^{-4}\text{ mol L}^{-1}\):
- Molar conductivity \(\Lambda_{m,2} = 147.1\text{ S cm}^2\text{ mol}^{-1}\).
- Calculate \(\sqrt{C_2}\):
\[ \sqrt{C_2} = \sqrt{9 \times 10^{-4}} = 3 \times 10^{-2} = 0.03 \]
- Substitute into the equation:
\[ 147.1 = \Lambda_m^\circ - A(0.03) \quad \text{--- (Equation 2)} \]
3. Solving the system of equations:
- Subtract Equation 2 from Equation 1:
\[ 149.1 - 147.1 = [\Lambda_m^\circ - A(0.01)] - [\Lambda_m^\circ - A(0.03)] \]
\[ 2.0 = A(0.03 - 0.01) \]
\[ 2.0 = A(0.02) \]
\[ A = \frac{2.0}{0.02} = 100 \]
- Substitute the value of \(A = 100\) back into Equation 1 to find \(\Lambda_m^\circ\):
\[ 149.1 = \Lambda_m^\circ - 100(0.01) \]
\[ 149.1 = \Lambda_m^\circ - 1.0 \]
\[ \Lambda_m^\circ = 149.1 + 1.0 = 150.1\text{ S cm}^2\text{ mol}^{-1} \]

Step 4: Final Answer:
The limiting molar conductivity of the \(\text{KCl}\) solution is \(150.1\text{ S cm}^2\text{ mol}^{-1}\), which corresponds to option (A).