What will be the number of cation vacancies if `10^(-3)` mol% of `CdCl_(2)` is added to the crystal of AgCl?

Explanation

When `10^(-3)` mol% of `CdCl_(2)` is added to the crystal of AgCl, `2xx10^(-3)`mol% of `Ag^(+)` ions are removed from the crystal lattice. As a result `10^(-3)`mol% of cation vacancies are generated. As in `CdCl_(2)`, cadmium exists as divalent cation `(Cd^(2+))`.
Thus, number of cation vacancies
`=10^(-3)mol%-=6.022xx10^(23)xx10^(-3)xx10^(-2)`
`=6.022xx10^(18)`