Chemistry - Solid State Question with Solution | TestHub

For each$\mathrm{S}{\mathrm{r}}^{2+}$added , one cationic vacancy is created, because two$\mathrm{N}{\mathrm{a}}^{+}$ions are removed and only one$\mathrm{S}{\mathrm{r}}^{2+}$is added.
So Number of cation vacancy$=2\times 10^{-3}\times 1$mol % of NaCl
$=\frac{2\times 10^{-3}}{100}$$\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$of NaCl
Total cationic vacancies$=2\times 10^{-5}\times 6.02\times 10^{23}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$of NaCl
$=12.04\times 10^{18}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$of NaCl