Question

molar ionic conductivities of a divalent cation and anion are \(57 \, \text{S cm}^2 \text{mol}^{-1}\) and \(73 \, \text{S cm}^2 \text{mol}^{-1}\), respectively. The molar conductivity of the solution of an electrolyte with the above cation and anion will be:

• A. \(65 \, \text{S cm}^2 \text{mol}^{-1}\)
• B. \(130 \, \text{S cm}^2 \text{mol}^{-1}\)
• C. \(187 \, \text{S cm}^2 \text{mol}^{-1}\)
• D. \(260 \, \text{S cm}^2 \text{mol}^{-1}\)

Answer 1

The Correct Option is B

The molar conductivity of a solution (Λ_Solution) is given by the sum of molar ionic conductivities:

Λ⁺²_C = 57 S cm² mol^-1
Λ⁺²_A = 73 S cm² mol^-1

Λ_Solution = λ⁺ + λ^-
Λ_Solution = 57 + 73 = 130 S cm² mol^-1

Answer 2

Step 1: Write the given data
Given molar ionic conductivities:
Cation (divalent): \( \lambda^+ = 57 \, \text{S cm}^2 \text{mol}^{-1} \)
Anion (divalent): \( \lambda^- = 73 \, \text{S cm}^2 \text{mol}^{-1} \)

Step 2: Identify the type of electrolyte
Since both ions are divalent (charges ±2), the simplest possible electrolyte will have the formula \( M X \), where \( M^{2+} \) and \( X^{2-} \) combine in a 1:1 ratio.

Step 3: Formula for molar conductivity
The molar conductivity at infinite dilution (\( \Lambda_m^0 \)) of an electrolyte is the sum of ionic conductivities of one formula unit of the compound:
\[ \Lambda_m^0 = \lambda^+ + \lambda^- \] because each formula unit produces one cation and one anion in solution.

Step 4: Substitute the values
\[ \Lambda_m^0 = 57 + 73 = 130 \, \text{S cm}^2 \text{mol}^{-1} \]

Step 5: Explanation for not multiplying by 2
The factor of ionic charge (±2) affects the magnitude of current each ion carries, but molar ionic conductivity values (\( \lambda^+ \) and \( \lambda^- \)) are already defined per mole of ions, taking their charge into account. Thus, we directly add the cationic and anionic values once for each ion produced per formula unit.

Final answer

\(130 \, \text{S cm}^2 \text{mol}^{-1}\)