Question

The molar conductivity of \(0.02 \text{ M HCl}\) is \(408.4 \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}\) at \(25^\circ\text{C}\). Calculate it's conductivity.

• A. \(16.180 \times 10^{-3} \Omega^{-1} \text{ cm}^{-1}\)
• B. \(8.168 \times 10^{-3} \Omega^{-1} \text{ cm}^{-1}\)
• C. \(6.108 \times 10^{-3} \Omega^{-1} \text{ cm}^{-1}\)
• D. \(4.042 \times 10^{-3} \Omega^{-1} \text{ cm}^{-1}\)

Hint:

Formula: \(\kappa = \frac{\Lambda_m \cdot C}{1000}\)

Answer 1

The Correct Option is B

Concept:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \] \[ \kappa = \frac{\Lambda_m \times C}{1000} \]

Step 1: Substitute values. \[ \kappa = \frac{408.4 \times 0.02}{1000} \]

Step 2: Calculate. \[ \kappa = \frac{8.168}{1000} = 8.168 \times 10^{-3} \Omega^{-1} \text{ cm}^{-1} \]

Step 3: Conclusion. \[ \text{Conductivity = } 8.168 \times 10^{-3} \Omega^{-1} \text{ cm}^{-1} \]