Question

The molar ionic conductivities of Ca²⁺ and Cl⁻ are 119.0 and 76.3 S cm⁻¹ mol⁻¹ respectively. The value of limiting molar conductivity of CaCl₂ will be:

• A. 195.3 S cm$^{-1}$ mol$^{-1}$
• B. 43.3 S cm$^{-1}$ mol$^{-1}$
• C. 314.3 S cm$^{-1}$ mol$^{-1}$
• D. 271.6 S cm$^{-1}$ mol$^{-1}$

Hint:

The limiting molar conductivity of an ionic compound is the sum of the molar ionic conductivities of the constituent ions.

Answer 1

The Correct Option is D

Step 1: The limiting molar conductivity of CaCl$_2$ is the sum of the molar ionic conductivities of its ions: \[ \lambda_{\text{m}}(\text{CaCl}_2) = \lambda_{\text{m}}(\text{Ca}^{2+}) + \lambda_{\text{m}}(\text{Cl}^{-}) \] Given: \[ \lambda_{\text{m}}(\text{Ca}^{2+}) = 119.0 \, \text{S cm}^{-1} \text{mol}^{-1}, \quad \lambda_{\text{m}}(\text{Cl}^{-}) = 76.3 \, \text{S cm}^{-1} \text{mol}^{-1} \] \[ \lambda_{\text{m}}(\text{CaCl}_2) = 119.0 + 2 \times 76.3 = 271.6 \, \text{S cm}^{-1} \text{mol}^{-1} \] Step 2: Thus, the correct limiting molar conductivity value for CaCl$_2$ is 271.6 S cm$^{-1}$ mol$^{-1}$.