If `NaCl` is doped with `10^(-4)"mol"%` of `srCl_(2)`, the concentration of cation vacancies will be `(N_(A)=6.023xx10^(23)"mol"^(-1))`

Explanation

Doping of NaCl with `10^(-3)` mol % `SrCl_(2)`, means that 100 moles of NaCl are doped with `10^(-3)` mol of `SrCl_(2)`.
1 mole of NaCl is doped with `SrCl_(2)= (10^(-3))/(100) "mole" = 10^(-5)` mole
As each `Sr^(2+)` ion introduces one cation vacancy, therefore, concentration of cation
vacancies = `10^(5)` mol/mol of NaCl = `10^(5)xx 6.02 xx 10^(23) "mol"^(-1) = 6.02 xx 10^(18) "mol"^(1)`