When NaCl is doped with $10^{-5}$ mol% $\mathrm{S}\mathrm{r}\mathrm{C}{\mathrm{l}}_2$ then what is the number of cationic vacancies per mole of NaCl?

Each molecule of $\mathrm{S}\mathrm{r}\mathrm{C}{\mathrm{l}}_2$ develops a cation vacancy hence number of cation vacancy per mole NaCl
Total number of molecules $=\mathrm{n}\times {\mathrm{N}}_{\mathrm{A}}$
$=\frac{10^{-5}}{100}\times {\mathrm{N}}_{\mathrm{A}}$
Where ${\mathrm{N}}_{\mathrm{A}}=6.023\times 10^{23}$
Total cationic vacancies = total molecules $\times$ vacancies created by 1 molecule
$=\frac{10^{-5}}{100}\times {\mathrm{N}}_{\mathrm{A}}\times 1$
$=10^{-7}{\mathrm{N}}_{\mathrm{A}}$