Doping of NaCl with `10^(-3)` mol % `SrCl_2` means that 100 moles of NaCl are doped with `10^(-3)` mol of `SrCl_2`
`therefore` 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)xx6.02xx10^23 "mol"^(-1) =6.02xx10^18 "mol"^(-1)`
