Question

The resistance of 0.05 M CH\(_3\)COOH solution is found to be 100 ohm. If the cell constant is 0.0354 cm\(^{-1}\), calculate the molar conductivity of the acetic acid solution.

Hint:

- The molar conductivity is calculated by dividing the conductivity by the molarity and multiplying by 1000. - The cell constant \(G\) is used to calculate the conductivity of the solution from the resistance.

Answer 1

The molar conductivity (\(\Lambda_m\)) can be calculated using the formula: \[ \Lambda_m = \frac{k}{M} \times 1000 \] Where: - \(k\) is the conductivity, - \(M\) is the molarity of the solution. First, calculate the conductivity (\(k\)) using the given values: \[ k = \frac{1}{R} \times L/A = \frac{G}{R} \] Where: - \(R = 100 \, \Omega\), - \(G = 0.0354 \, \text{cm}^{-1}\). Thus: \[ k = \frac{1}{100} \times 0.0354 = 3.54 \times 10^{-4} \, \Omega^{-1} \, \text{cm}^{-1}. \] Now, calculate \(\Lambda_m\) using the formula: \[ \Lambda_m = \frac{3.54 \times 10^{-4}}{0.05} \times 1000 = 7.08 \, \text{S cm}^2 \, \text{mol}^{-1}. \] The molar conductivity of the acetic acid solution is \(7.08 \, \text{S cm}^2 \, \text{mol}^{-1}\).