If NaCl is doped with `10^(-3)` mol% `SrCl_(2)`, what is the concentration of cation vacancies?

Explanation

If crystal lattice of NaCl is doped with `SrCl_(2)`, two `Na^(+)` ions are removed from their sites for each `Sr^(2+)` ion, resulting in two cationic vacancies. One of these sites is occupied by `Sr^(2+)` ion, while the other remains vacant.
According to the question, NaCl is doped with `10^(-3)` mol% of `SrCl_(2)`. This means 100 mol of NaCl are doped with `10^(-3)` mol of `SrCl_(2)`.
`10^(-3)` mol of `SrCl_(2)=6.023xx10^(23)xx10^(-3)=6.023xx10^(20) Sr^(2+)` ions. For each `Sr^(2+)` ion, one cationic vacancy is created in the crystal lattice of NaCl. Hence, number of cationic vacancies created when 100 mol of NaCl is doped with `10^(-3)` mol% of NaCl is `6.023xx10^(20)`. For 1 mole of NaCl, the number of cation vacancies is `6.023xx10^(18)`.