The `EMF` of the cell `:`
`Ag|AgCl,0.1 MKCl||0.1 M AgNO_(3)|Ag` is `0.45V. 0.1 M KCl `is `85%` dissociated and `0.1 M AgNO_(3)` is `82%` dissociated. Calculate the solubility product of `AgCl` at `25^(@)C`.

Explanation

To calculate the solubility product (Ksp) of AgCl at 25°C using the given data, we can follow these steps:

### Step 1: Determine the Concentrations of Ions
We are given the dissociation percentages of KCl and AgNO3, which will help us find the concentrations of Cl⁻ and Ag⁺ ions.

- For KCl (0.1 M, 85% dissociated):
\[
[Cl^-] = 0.1 \, \text{M} \times \frac{85}{100} = 0.085 \, \text{M}
...