Question

The conductivity of 0.02 M of KCl solution at 298 K is 0.0123 \(\text{ohm}^{-1} \text{cm}^{-1}\). If the resistance of the cell containing this solution is 120 ohms, what is the value of the cell constant ?

• A. \(1.050 \text{cm}^{-1}\)
• B. \(1.025 \text{cm}^{-1}\)
• C. \(1.376 \text{cm}^{-1}\)
• D. \(1.476 \text{cm}^{-1}\)

Hint:

Cell constant \(= \kappa \times R\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Cell constant is the ratio \(l/A\) of the conductivity cell. Conductivity is \(\kappa = \dfrac{1}{R}\cdot\dfrac{l}{A}\).

Step 2: Key Formula or Approach
\[ G^* = \frac{l}{A} = \kappa\,R \]

Step 3: Detailed Explanation
\[ G^* = 0.0123\ \text{ohm}^{-1}\text{cm}^{-1} \times 120\ \text{ohm} = 1.476\ \text{cm}^{-1} \]
The molarity 0.02 M is not needed for this step, because the conductivity already contains that information.

Final Answer:
The cell constant is 1.476 cm\(^{-1}\), option (D). \[ \boxed{1.476\ \text{cm}^{-1}\ \text{(D)}} \]