Question

Resistance of 2 M solution of an electrolyte is $50\text{ }\Omega$. The conductivity of the solution is $3\text{ S m}^{-1}$. If the resistance of 4 M solution of the same electrolyte is $260\text{ }\Omega$, its molar conductivity is:

• A. $6.25 \times 10^{-3}\text{ S m}^2\text{ mol}^{-1}$
• B. $6.25 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1}$
• C. $625 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1}$
• D. $62.5 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1}$

Hint:

To ensure your units track correctly in SI calculations, remember that converting concentration from $\text{mol/L}$ to $\text{mol/m}^3$ requires multiplying by $1000$. This places $1000$ in the denominator, yielding $\frac{0.25}{400} = 6.25 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1}$.

Answer 1

The Correct Option is B

Concept: Conductivity ($\kappa$) is related to resistance ($R$) by the cell constant ($G^*$): $\kappa = \frac{G^*}{R}$. The cell constant depends solely on the physical geometry of the cell electrodes, meaning its value remains fixed for a given cell. Molar conductivity ($\Lambda_m$) is calculated from conductivity and molar concentration ($C$) using: \[ \Lambda_m = \frac{\kappa}{1000 \times C} \quad (\text{when using SI units of S m}^2\text{ mol}^{-1}) \]

Step 1: Calculate the cell constant ($G^*$) using the first solution data.
Given $C_1 = 0.2\text{ M}$, $R_1 = 50\text{ }\Omega$, and $\kappa_1 = 1.3\text{ S m}^{-1}$: \[ G^* = \kappa_1 \times R_1 = 1.3 \times 50 = 65\text{ m}^{-1} \]

Step 2: Calculate conductivity ($\kappa_2$) for the second solution.
Given $C_2 = 0.4\text{ M}$ and $R_2 = 260\text{ }\Omega$: \[ \kappa_2 = \frac{G^*}{R_2} = \frac{65}{260} = 0.25\text{ S m}^{-1} \]

Step 3: Calculate the molar conductivity ($\Lambda_m$) in SI units.
Using $C_2 = 0.4\text{ mol L}^{-1} = 0.4 \times 10^3\text{ mol m}^{-3}$: \[ \Lambda_m = \frac{\kappa_2}{1000 \times C_2} = \frac{0.25}{1000 \times 0.4} = \frac{0.25}{400} \] \[ \Lambda_m = 0.000625 = 6.25 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1} \]