Resistance of a conductivity cell filled with 0.1 mol `L^(-1)` KCl solution is `100Omega`. If the resistance of the same cell when filled with 0.02 mol `L^(-1)` KCl solution is `520Omega`, calculate the conductivity and molar conductivity of 0.02 M KCl solution. The conductivity of 0.1 M KCl solution is 1.29S/m.

Explanation

To solve the problem, we need to calculate the conductivity and molar conductivity of the 0.02 M KCl solution using the given data. Here’s a step-by-step solution:

### Step 1: Understand the relationship between resistance, conductivity, and molar conductivity
The conductivity (κ) of a solution is related to the resistance (R) of the cell and the cell constant (K) by the formula:
\[ \kappa = \frac{K}{R} \]
Where:
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