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. What is the conductivity of 0.02 M KCl solution ?

Options

A. 0.452 S `m^(-1)`
B. 0.215 S `m^(-1)`
C. 0.614 S `m^(-1)`
D. 0.433 S `m^(-1)`

Explanation

To find the conductivity of the 0.02 M KCl solution, we can use the relationship between conductivity (κ), conductance (G), and the cell constant (K) given by the formula:

\[ \kappa = G \cdot K \]

### Step 1: Calculate the conductance of the 0.1 M KCl solution
The conductance (G) is the reciprocal of resistance (R). For the 0.1 M KCl solution, the resistance is given as 200 Ohm.

\[ G_{0.1} = \frac{1}{R_{0.1}} = \frac{1}{200} \, \text{S} \]

### Step 2: Use the known conductivity to find the cell constant
We know the conductivity of the 0.1 M KCl solution is 1.29 S/m. Using the formula:

\[ \kappa_{0.1} = G_{0.1} \cdot K \]

Substituting the values:

\[ 1.29 = \left(\frac{1}{200}\right) \cdot K \]

To find the cell constant (K), rearranging gives:

\[ K = 1.29 \cdot 200 = 258 \, \text{m}^{-1} \]

### Step 3: Calculate the conductance of the 0.02 M KCl solution
For the 0.02 M KCl solution, the resistance is given as 420 Ohm.

\[ G_{0.02} = \frac{1}{R_{0.02}} = \frac{1}{420} \, \text{S} \]

### Step 4: Calculate the conductivity of the 0.02 M KCl solution
Using the cell constant we found earlier, we can now find the conductivity of the 0.02 M KCl solution:

\[ \kappa_{0.02} = G_{0.02} \cdot K \]

Substituting the values:

\[ \kappa_{0.02} = \left(\frac{1}{420}\right) \cdot 258 \]

Calculating this gives:

\[ \kappa_{0.02} = \frac{258}{420} \approx 0.614 \, \text{S/m} \]

### Final Answer
The conductivity of the 0.02 M KCl solution is approximately **0.614 S/m**.

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