If NaCl is doped with `10^(-3)` mol% of `SrCI_2` what is the concentration of cation vacancies?

Explanation

As NaCl is doped with `10^(-3)` mol % of `SrCI_2` It means 100 mol of NaCl are dopped with `10^(-3)` mol of `SrCI_2` I mo! of NaCl is dopped with `srCI_2=(10^(-3))/100mol=10^(-5)mol`
Concentration of cation vacancies `=10^(-5)xx6.02xx10^(23)mol^(-1)=6.02xx10^(18)mol^(-1)`