If NaCl is doped with 10^-4 mol % of SrCl₂, the concentration of cation vacancies will be $\left({\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{ \mathrm{mol}}^{-1}\right)$

Moles of SrCl₂​ in 1 mole NaCl=$\frac{{10}^{-4}}{100}={10}^{-6}\mathrm{moles}$

One Sr⁺² creates one cationic vacancies.

Concentration of the cationic vacancies produced by 10⁻⁶ moles of SrCl_2​

=10⁻⁶×6.023×10²³                                            
=6.023×10¹⁷ mol⁻¹