Question

Explain molar conductivity. The resistance of a cell filled with 0.02 mol L$^{-1$ KCl solution is 480 $\Omega$. Calculate the molar conductivity of the solution. (Cell constant = 1.29 cm$^{-1}$).}

Hint:

\(\kappa = (\text{cell constant})\times G\) with cell constant in cm$^{-1}$ and \(G\) in S.
Use \(\Lambda_m = \kappa \dfrac{1000}{c}\) (with \(c\) in mol L$^{-1}$) to get \(\text{S cm}^2\text{ mol}^{-1}\).

Answer 1

Definition:
Molar conductivity \((\Lambda_m)\) is the conductance of all the ions produced by 1 mole of electrolyte placed between two electrodes 1 cm apart, with the solution having unit cross-sectional area.
It is given by \(\Lambda_m = \kappa \dfrac{1000}{c}\), where \(\kappa\) is conductivity (S cm$^{-1}$) and \(c\) is concentration in mol L$^{-1}$.

Step 1: Find conductivity \(\kappa\).
Conductance \(G = \dfrac{1}{R} = \dfrac{1}{480}\ \text{S} = 2.0833\times 10^{-3}\ \text{S}\).
\(\kappa = (\text{cell constant}) \times G = 1.29\ \text{cm}^{-1} \times 2.0833\times 10^{-3}\ \text{S} = 2.6875\times 10^{-3}\ \text{S cm}^{-1}.\)

Step 2: Calculate molar conductivity.
\(c = 0.02\ \text{mol L}^{-1}\).
\[ \Lambda_m = \kappa \frac{1000}{c} = 2.6875\times 10^{-3}\ \frac{1000}{0.02} = 2.6875\times 10^{-3}\times 5.0\times 10^{4} = 1.34375\times 10^{2}\ \text{S cm}^2\text{ mol}^{-1}. \]
\[ \boxed{\Lambda_m \approx 1.34\times 10^{2}\ \text{S cm}^2\ \text{mol}^{-1}} \]