When NaCl is dopped with `10^(-5) " mole % of" SrCl_(2)`, what is the no. of cationic vacanies?

Explanation

Doping of NaCl with `10^(-3)` mol% `SrCl_(2)` means that 100 mol of NaCl are droped with `10^(-3)` mol of `SrCl_(2)`.
`therefore` 1 mol of NaCl is doped with `=10^(-3)/(100)` mol
`=10^(-5)` mol
As each `Sr^(2+)` ion introduces are cation vacancy, therefore, concentration of cation vacancies
`=10^(-5)` mol/mol of NaCl
`=10^(-5)xx6.022xx10^(23)"mol"^(-1)` of NaCl
`=6.02xx10^(18)"mol"^(-1)` of NaCl