Question

The resistance of a conductivity cell filled with 0.01N KC/ at 25°C was found to be 5002. The specific conductance of 0.01 N KC/ at 25°C is 1.41×10³,¯¹cm¯¹. The resistance of same cell filled with 0.3 NZnSO, at 25°C was found to be 692. The equivalent conductivity of ZnSO, solution is

Answer 1

34.06 $\Omega^{-1}cm^2 \text{eq}^{-1}$

Answer 2

1. Calculate the cell constant ($G^*$) using the given resistance ($R_{KCl}$) and specific conductance ($\kappa_{KCl}$) of the $KCl$ solution: $G^* = \kappa_{KCl} \times R_{KCl}$.
2. Calculate the specific conductance ($\kappa_{ZnSO_4}$) of the $ZnSO_4$ solution using the cell constant and its resistance ($R_{ZnSO_4}$): $\kappa_{ZnSO_4} = G^* / R_{ZnSO_4}$.
3. Calculate the equivalent conductivity ($\Lambda_e$) of the $ZnSO_4$ solution using its specific conductance and normality ($N_{ZnSO_4}$): $\Lambda_e = \kappa_{ZnSO_4} \times (1000/N_{ZnSO_4})$.

Given: $R_{KCl} = 500 \Omega$ $\kappa_{KCl} = 1.41 \times 10^{-3} \Omega^{-1}cm^{-1}$ $R_{ZnSO_4} = 69 \Omega$ $N_{ZnSO_4} = 0.3 N$

Step 1: $G^* = (1.41 \times 10^{-3} \Omega^{-1}cm^{-1}) \times (500 \Omega) = 0.705 cm^{-1}$ Step 2: $\kappa_{ZnSO_4} = 0.705 cm^{-1} / 69 \Omega \approx 0.010217 \Omega^{-1}cm^{-1}$ Step 3: $\Lambda_e = (0.010217 \Omega^{-1}cm^{-1}) \times (1000/0.3) \approx 34.06 \Omega^{-1}cm^2 \text{eq}^{-1}$