Question

The conductivity of centimolar solution of KCl at 298 K is 0.021 $\Omega^{-1}$ cm$^{-1}$. The resistance of the cell containing the solution at 298 K is 60 $\Omega$. The value of cell constant is

• A. 3.28 cm$^{-1}$
• B. 1.26 cm$^{-1}$
• C. 3.54 cm$^{-1}$
• D. 1.34 cm$^{-1}$

Hint:

The cell constant is a physical property of the conductivity cell and remains constant for a given cell, regardless of the solution placed inside it or the temperature. It is often determined using a solution of known conductivity (like KCl).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question provides the conductivity ($\kappa$) and resistance ($R$) of a centimolar KCl solution in a conductivity cell at a specific temperature. We need to calculate the cell constant ($G^*$) of the conductivity cell.
Step 2: Key Formula or Approach:
The relationship between conductivity, resistance, and cell constant is given by:
\[ \kappa = \frac{1}{R} \times G^* \]
Where:
- \(\kappa\) is the conductivity (units: $\Omega^{-1}$ cm$^{-1}$ or S cm$^{-1}$).
- \(R\) is the resistance (units: $\Omega$).
- \(G^*\) is the cell constant (units: cm$^{-1}$).
Rearranging the formula to solve for the cell constant:
\[ G^* = \kappa \times R \]
Step 3: Detailed Explanation:
Given values:
- Conductivity, \( \kappa = 0.021 \ \Omega^{-1} \text{ cm}^{-1} \)
- Resistance, \( R = 60 \ \Omega \)
Substitute these values into the formula for cell constant:
\[ G^* = (0.021 \ \Omega^{-1} \text{ cm}^{-1}) \times (60 \ \Omega) \]
\[ G^* = 1.26 \text{ cm}^{-1} \]
The concentration ("centimolar") and temperature (298 K) are relevant for conductivity but not directly used in calculating the cell constant, as the cell constant is a geometric property of the cell itself, independent of the solution.
Step 4: Final Answer:
The value of the cell constant is 1.26 cm$^{-1}$.