Question

The relation between molar conductivity \( \Lambda_m \), conductivity \( \kappa \), and concentration \( C \) of an electrolyte solution is:

• A. \( \Lambda_m = \dfrac{\kappa}{1000C} \)
• B. \( \Lambda_m = \dfrac{C}{\kappa \times 1000} \)
• C. \( \Lambda_m = \dfrac{\kappa \times 1000}{C} \)
• D. \( \Lambda_m = \kappa C \)

Hint:

Remember the key electrochemistry relation: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] On dilution, concentration \(C\) decreases and therefore molar conductivity \( \Lambda_m \) increases.

Answer 1

The Correct Option is C

Concept: Molar conductivity (\( \Lambda_m \)) is defined as the conductivity of the solution containing one mole of electrolyte. It is related to conductivity \( (\kappa) \) and concentration \( (C) \) by the relation: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] where

• \( \Lambda_m \) = Molar conductivity \( (\text{S cm}^2 \text{ mol}^{-1}) \)
• \( \kappa \) = Conductivity of solution \( (\text{S cm}^{-1}) \)
• \( C \) = Concentration of electrolyte in \( \text{mol L}^{-1} \)

The factor 1000 is used to convert litres to \( \text{cm}^3 \) since \(1\,L = 1000\,cm^3\).
Step 1: {Write the definition of molar conductivity.} \[ \Lambda_m = \kappa \times V \] where \(V\) is the volume containing one mole of electrolyte.
Step 2: {Express volume in terms of concentration.} If the concentration of solution is \(C\) mol L\(^{-1}\), \[ V = \frac{1}{C} \, L \] Converting litres into \(cm^3\): \[ V = \frac{1000}{C} \, cm^3 \]
Step 3: {Substitute the value of volume in the equation.} \[ \Lambda_m = \kappa \times \frac{1000}{C} \] \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] Thus, the correct relation is option (C).