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

Explanation

To solve the problem, we need to calculate the conductivity and molar conductivity of a 0.02 M KCl solution based on the given data. Let's break it down step by step.

### Step 1: Calculate the Cell Constant (G*)
We are given the resistance (R1) of a 0.1 M KCl solution and its conductivity (k1). The formula for conductivity is:

\[ k = \frac{G^*}{R} \]
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