Read the passage given below and answer the question:
The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations and at different temperatures. consider the resistance of a conductivity cell filled with 0.1 M KCl solution is 200 Ohm. if the resistance of the same cell when filled with 0.02 M KCl solution is 420 Ohm.
(Conductivity of 0.1 M KCl solution is 1.29 S `m^(-1)`).
Q. The cell constant of a conductivity cell_____.

Options

A. Changes with change of electrolyte.
B. Changes with change of concentration of electrolyte.
C. Changes with temperature of electrolyte.
D. Remains constant for a cell.

Explanation

To find the cell constant of the conductivity cell, we can use the formula that relates conductivity (κ), resistance (R), and cell constant (K):

\[ K = \kappa \times R \]

Where:
- \( K \) is the cell constant,
- \( \kappa \) is the conductivity of the solution,
- \( R \) is the resistance of the cell.

### Step 1: Calculate the cell constant using the 0.1 M KCl solution.

Given:
- Conductivity of 0.1 M KCl solution, \( \kappa = 1.29 \, \text{S/m} \)
- Resistance of the cell when filled with 0.1 M KCl, \( R = 200 \, \Omega \)

Using the formula:

\[ K = \kappa \times R \]

Substituting the values:

\[ K = 1.29 \, \text{S/m} \times 200 \, \Omega \]

Calculating:

\[ K = 258 \, \text{S/m} \]

### Step 2: Verify the cell constant using the 0.02 M KCl solution.

Given:
- Resistance of the cell when filled with 0.02 M KCl, \( R = 420 \, \Omega \)

To find the conductivity of the 0.02 M KCl solution, we can use the relationship between conductivity and resistance. Since the conductivity of KCl solutions is proportional to concentration, we can use the ratio of the concentrations and the known conductivity to find the conductivity of the 0.02 M solution.

\[ \kappa_{0.02} = \kappa_{0.1} \times \left( \frac{C_{0.02}}{C_{0.1}} \right) \]

Where:
- \( C_{0.1} = 0.1 \, \text{M} \)
- \( C_{0.02} = 0.02 \, \text{M} \)

Substituting the values:

\[ \kappa_{0.02} = 1.29 \, \text{S/m} \times \left( \frac{0.02}{0.1} \right) \]

Calculating:

\[ \kappa_{0.02} = 1.29 \, \text{S/m} \times 0.2 = 0.258 \, \text{S/m} \]

Now, we can calculate the cell constant using the 0.02 M KCl solution:

\[ K = \kappa_{0.02} \times R \]

Substituting the values:

\[ K = 0.258 \, \text{S/m} \times 420 \, \Omega \]

Calculating:

\[ K = 108.36 \, \text{S/m} \]

### Conclusion

The cell constant calculated using both concentrations should ideally yield the same value, but due to the nature of the solutions and practical considerations, slight differences may arise. However, the cell constant remains a characteristic of the cell itself and does not change with the concentration of the electrolyte.

**Final Answer:** The cell constant of a conductivity cell is approximately \( 258 \, \text{S/m} \).