A conductivity cell when filled with 0.01 M KCl has a resistance of `745 Omega` at `25^(@) C`. When the same cell was filled with an aqueous solution of 0.005 M `CaCl_(2)` solution the resistance was `874 Omega` . Calculate
(i) Conductivity of solution
(ii) Molar conductivity of solution .
[Conductivity of 0.01 M KCl = 0.141 ` S m^(-1)`]

Explanation

To solve the problem step by step, we will first calculate the cell constant using the KCl solution, then use that to find the conductivity of the CaCl₂ solution, and finally calculate the molar conductivity.

### Step 1: Calculate the Cell Constant (G*)
The cell constant (G*) can be calculated using the conductivity (κ) of the KCl solution and its resistance (R).

The formula for cell constant is:
\[
G^* = \kappa \times R
...