Question

A conductivity cell is filled with a solution of KCl of concentration 0.2 mol dm\(^{-3}\) and its conductivity is 0.28 S m\(^{-1}\). The resistance of this solution is 82.2 Ω. The same cell filled with a solution of 0.0025 mol dm\(^{-3}\) K₂SO₄ showed a resistance of 325 Ω. The molar conductivity of K₂SO₄ solution (in S m² mol\(^{-1}\)) is: 
 

• A. \(1.4 \times 10^{-2}\)
• B. \(2.8 \times 10^{-2}\)
• C. \(4.2 \times 10^{-3}\)
• D. \(5.6 \times 10^{-4}\)

Hint:

When solving conductivity problems, always determine the cell constant first using a standard solution, and then use it to calculate the conductivity of the unknown solution.

Answer 1

The Correct Option is B

The relationship between conductivity (\(\kappa\)), resistance (\(R\)), and cell constant (\(G^*\)) is given by: \[ \kappa = \frac{G^*}{R} \] First, we calculate the cell constant (\(G^*\)) using the KCl solution: \[ G^* = \kappa_{{KCl}} \times R_{{KCl}} = 0.28 \times 82.2 = 23.016 { m}^{-1} \] Now, using the resistance of the K₂SO₄ solution: \[ \kappa_{{K}_2{SO}_4} = \frac{G^*}{R_{{K}_2{SO}_4}} = \frac{23.016}{325} = 0.0708 { S m}^{-1} \] Molar conductivity (\(\Lambda_m\)) is given by: \[ \Lambda_m = \frac{\kappa}{C} \] Substituting the values: \[ \Lambda_m = \frac{0.0708}{0.0025} = 28.32 { S cm}^2 { mol}^{-1} \] Since \(1 { S cm}^2 { mol}^{-1} = 10^{-4} { S m}^2 { mol}^{-1}\), converting: \[ \Lambda_m = 2.8 \times 10^{-2} { S m}^2 { mol}^{-1} \] Thus, the molar conductivity of K₂SO₄ solution is \(2.8 \times 10^{-2}\) S m² mol\(^{-1}\).