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