Question

If resistivity of 0.8 M KCl solution is \( 2.5 \times 10^{-3} \, \Omega \, \text{cm} \), calculate molar conductivity of solution?

• A. \( 3 \times 10^{5} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
• B. \( 2 \times 10^{5} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
• C. \( 4 \times 10^{5} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
• D. \( 5 \times 10^{5} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)

Hint:

To calculate molar conductivity, use the formula \( \Lambda_m = \frac{1}{\rho} \times 1000 \times C \), where \( \rho \) is resistivity and \( C \) is concentration.

Answer 1

The Correct Option is D

Step 1: Formula for molar conductivity.
Molar conductivity \( \Lambda_m \) is related to resistivity \( \rho \) by the formula: \[ \Lambda_m = \frac{1}{\rho} \times 1000 \times C \] Where: - \( C \) is the concentration in mol/L.

Step 2: Calculation.
Given: - \( \rho = 2.5 \times 10^{-3} \, \Omega \, \text{cm} \), - \( C = 0.8 \, \text{mol/L} \), Substitute the values into the formula: \[ \Lambda_m = \frac{1}{2.5 \times 10^{-3}} \times 1000 \times 0.8 = 5 \times 10^{5} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]
Step 3: Conclusion.
The molar conductivity of the solution is \( 5 \times 10^{5} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \), so the correct answer is (D) \( 5 \times 10^{5} \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \).