Question

When a certain conductivity cell was filled with 0.05 M KCl solution, it has a resistance of 100 ohms at 25°C. When the same cell was filled with 0.02 M $\text{AgNO_{3}$ solution, the resistance was 90 ohms. Calculate the conductivity and molar conductivity of $\text{AgNO}_{3}$ solution.} % Given: \text{Conductivity of 0.05 M KCl solution = \( 1.35 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{-1} \)}

Hint:

To calculate molar conductivity, the formula \( \Lambda_m = \text{Conductivity} \times \frac{1000}{C} \) is used, where \( C \) is the concentration of the solution in mol/L.

Answer 1

To calculate the conductivity of $\text{AgNO}_{3}$ solution, we will use the relationship between the conductivity, resistance, and the cell constant. The cell constant (G*) is calculated using the following equation: \[ \text{Cell constant (G*)} = \text{Conductivity} \times \text{Resistance} \] For KCl solution: \[ G^* = 1.35 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{-1} \times 100 \, \Omega = 1.35 \, \text{cm}^{-1} \] Now, for the $\text{AgNO}_{3}$ solution, we can use the same cell constant: \[ G^* = 1.35 \, \text{cm}^{-1} = \text{Conductivity} \times 90 \, \Omega \] \[ \text{Conductivity} = \frac{1.35}{90} = 0.015 \, \text{S/cm} \] Next, we calculate the molar conductivity ($\Lambda_m$) using the formula: \[ \Lambda_m = \text{Conductivity} \times \frac{1000}{\text{Concentration}} \] For $\text{AgNO}_{3}$ solution with concentration \( 0.02 \, \text{M} \): \[ \Lambda_m = 0.015 \, \text{S/cm} \times \frac{1000}{0.02} = 750 \, \text{Scm}^2/\text{mol} \] Thus, the molar conductivity of the $\text{AgNO}_{3}$ solution is \( 750 \, \text{Scm}^2/\text{mol} \).