Question

Calculate the molar conductance of 0.025M aqueous solution of calcium chloride at 25°C. The specific conductance of calcium chloride is \( 12.04 \times 10^{-2} \, \text{S} \, \text{m}^{-1} \)

• A. \( 4.816 \times 10^{-5} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)
• B. \( 3.816 \times 10^{-5} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)
• C. \( 3.816 \times 10^{-5} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)
• D. \( 481.6 \times 10^{-5} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)

Hint:

Molar conductance can be calculated using the specific conductance and the concentration of the solution.

Answer 1

The Correct Option is D

Molar conductance \( \Lambda_m \) is given by the formula: \[ \Lambda_m = \kappa \cdot \frac{1}{c} \] Where: - \( \kappa \) is the specific conductance (\( 12.04 \times 10^{-2} \, \text{S} \, \text{m}^{-1} \)) - \( c \) is the concentration of the solution (0.025M) Substitute the values into the equation: \[ \Lambda_m = \frac{12.04 \times 10^{-2}}{0.025} = 481.6 \times 10^{-5} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \] Thus, the molar conductance is \( 481.6 \times 10^{-5} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \).