Question

The formula used to calculate molar conductivity of an electrolyte is _______.

• A. \( \Lambda = \frac{1000c}{k} \)
• B. \( c = \frac{1000\Lambda}{k} \)
• C. \( \Lambda = \frac{1000k}{c} \)
• D. \( k = \frac{1000}{\Lambda c} \)

Hint:

The "1000" in the molar conductivity formula is a unit conversion factor, not a fundamental constant. It's used specifically when concentration is in mol/L and conductivity is in S/cm. Always be mindful of the units you are given and the units you need.

Answer 1

The Correct Option is C

Step 1: Define the terms.
- Molar conductivity (\(\Lambda_m\)): The conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.
- Conductivity (k or \(\kappa\)): The conductance of a solution of 1 cm length with a cross-sectional area of 1 cm\(^2\). Its unit is S cm\(^{-1}\).
- Concentration : The amount of electrolyte in moles per liter (mol L\(^{-1}\)).
Step 2: Relate the terms. Molar conductivity is defined as the conductivity divided by the molar concentration: \[ \Lambda_m = \frac{k}{c} \] However, the units must be consistent. Typically, \(k\) is in S cm\(^{-1}\) and \(c\) is in mol L\(^{-1}\). To make them compatible, we must convert the volume from Liters to cm\(^3\) (since 1 L = 1000 cm\(^3\)).
\[ \Lambda_m (\text{S cm}^2 \text{mol}^{-1}) = \frac{k (\text{S cm}^{-1})}{c (\text{mol L}^{-1})} \times \frac{1000 (\text{cm}^3)}{1 (\text{L})} \] \[ \Lambda_m = \frac{1000k}{c} \]