Doping of \(\mathrm{NaCl}\) with \(10^{-3} \mathrm{~mol} \% \mathrm{SrCl}_2\) means that 100 moles of \(\mathrm{NaCl}\) are doped with \(10^{-3} \mathrm{~mol}\) of \(\mathrm{SrCl}_2\).
\(\therefore 1\) mole of \(\mathrm{NaCl}\) is doped with \(\mathrm{SrCl}_2\)
\(=\frac{10^{-3}}{100} \text { mole }=10^{-5} \text { mole. }\)
Each \(\mathrm{Sr}^{2+}\) introduces one cation vacancy, therefore, concentration of cation vacancies
\(\begin{aligned}
&=10^{-5} \times \mathrm{N}_{\mathrm{A}} \\
&=10^{-5} \mathrm{~mol} \times 6.023 \times 10^{23} \mathrm{~mol}^{-1}=6.023 \times 10^{18} .
\end{aligned}\)