Question

When \( 0.1 \, mol \, L^{-1} \) of KCl was filled in a Conductivity cell the resistance was 80 ohms at 298 K. Conductivity of \( 0.1 \, M \) KCl at 298 K is \( 1.29 \, S \, m^{-1} \). The same cell when filled with an unknown electrolyte of concentration \( 0.025 \, M \), had a resistance of 92 ohms. What is the Molar conductivity of the electrolyte at the given concentration?

• A. \( 220.6 \, S \, cm^2 \, mol^{-1} \)
• B. \( 0.2206 \, S \, cm^2 \, mol^{-1} \)
• C. \( 448 \, S \, cm^2 \, mol^{-1} \)
• D. \( 0.449 \, S \, cm^2 \, mol^{-1} \)

Hint:

First calculate the cell constant using standard KCl solution, then use it to find conductivity of the unknown solution and finally apply \( \Lambda_m = \frac{\kappa \times 1000}{C} \).

Answer 1

The Correct Option is C

Step 1: Find cell constant using KCl solution.
Conductivity is given by:
\[ \kappa = \frac{\text{Cell constant}}{R} \]
Therefore:
\[ \text{Cell constant} = \kappa \times R \]
For KCl solution:
\[ \kappa = 1.29 \, S \, m^{-1} \]
\[ R = 80 \, \Omega \]
So:
\[ \text{Cell constant} = 1.29 \times 80 \]
\[ = 103.2 \, m^{-1} \]

Step 2: Find conductivity of unknown electrolyte.
For the unknown electrolyte:
\[ R = 92 \, \Omega \]
Using:
\[ \kappa = \frac{\text{Cell constant}}{R} \]
\[ \kappa = \frac{103.2}{92} \]
\[ \kappa = 1.1217 \, S \, m^{-1} \]

Step 3: Convert concentration into proper unit.
Given concentration:
\[ C = 0.025 \, mol \, L^{-1} \]
For molar conductivity in \( S \, cm^2 \, mol^{-1} \), use:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
Here conductivity should be in \( S \, cm^{-1} \).
Convert:
\[ 1.1217 \, S \, m^{-1} = 0.011217 \, S \, cm^{-1} \]

Step 4: Calculate molar conductivity.
\[ \Lambda_m = \frac{0.011217 \times 1000}{0.025} \]
\[ \Lambda_m = \frac{11.217}{0.025} \]
\[ \Lambda_m = 448.68 \, S \, cm^2 \, mol^{-1} \]

Step 5: Conclusion.
The molar conductivity of the electrolyte is approximately:
\[ 448 \, S \, cm^2 \, mol^{-1} \]
Therefore:
\[ \boxed{448 \, S \, cm^2 \, mol^{-1}} \]