Question

The resistance of a conductivity cell filled with 0.02 M KCl solution is 85 \(\Omega\) at \(25^{\circ}C\). Conductivity of this solution is \(0.3 \text{ S m}^{-1}\). Resistance of 0.0025 M \(K_2SO_4\) solution taken in the same cell is 300 \(\Omega\). The molar conductivity of 0.0025 M \(K_2SO_4\) solution (in S \(\text{m}^2 \text{mol}^{-1}\)) is:

• A. \(6.8 \times 10^{-3}\)
• B. \(2.4 \times 10^{-2}\)
• C. \(3.4 \times 10^{-2}\)
• D. \(3.4 \times 10^{-3}\)

Hint:

Remember the conversion factor: \(1 \text{ M} = 1000 \text{ mol/m}^3\).

Answer 1

The Correct Option is C

Concept: The cell constant (\(G^*\)) is determined as \(G^* = \kappa \times R\). Molar conductivity (\(\Lambda_m\)) is defined as \(\Lambda_m = \frac{\kappa}{C}\).

Step 1: Calculate the cell constant using KCl.
\[ G^* = 0.3 \text{ S m}^{-1} \times 85 \text{ } \Omega = 25.5 \text{ m}^{-1} \]

Step 2: Calculate conductivity (\(\kappa\)) of \(K_2SO_4\).
\[ \kappa_{K_2SO_4} = \frac{G^*}{R} = \frac{25.5}{300} = 0.085 \text{ S m}^{-1} \]

Step 3: Calculate molar conductivity.
Concentration \(C = 0.0025 \text{ mol/L} = 0.0025 \times 1000 \text{ mol/m}^3 = 2.5 \text{ mol/m}^3\). \[ \Lambda_m = \frac{0.085}{2.5} = 0.034 \text{ S m}^2 \text{ mol}^{-1} \] Adjusting the decimal for the option format: \[ \Lambda_m = 3.4 \times 10^{-2} \text{ S m}^2 \text{ mol}^{-1} \] \[ \boxed{3.4 \times 10^{-2}} \]