Question

The conductivity of 0.02 mol L\(^{-1}\) KCl solution is 0.248 S m\(^{-1}\). Its molar conductivity is

• A. 20 S m\(^2\) mol\(^{-1}\)
• B. \( 1.24 \times 10^{-3} \) S m\(^2\) mol\(^{-1}\)
• C. \( 1.24 \times 10^{-4} \) S m\(^2\) mol\(^{-1}\)
• D. \( 2.48 \times 10^{-2} \) S m\(^2\) mol\(^{-1}\)
• E. \( 1.24 \times 10^{-2} \) S m\(^2\) mol\(^{-1}\)

Hint:

To calculate molar conductivity, divide the conductivity by the concentration of the solution.

Answer 1

Answer: (E)

Molar conductivity \( \lambda \) is given by: \[ \lambda = \frac{\kappa}{C} \] where \( \kappa \) is the conductivity and \( C \) is the concentration.
Given: 
- \( \kappa = 0.248 \, {S m}^{-1} \), 
- \( C = 0.02 \, {mol L}^{-1} \). 
Thus: \[ \lambda = \frac{0.248}{0.02} = 1.24 \times 10^{-2} \, {S m}^2 {mol}^{-1} \] Thus, the correct answer is (E).