Chemistry - Solid State Question with Solution | TestHub

As each$\mathrm{S}{\mathrm{r}}^{2+}$ion introduces one cation vacancy, therefore, concentration of cation vacancies = mole% of$\mathrm{S}\mathrm{r}\mathrm{C}{\mathrm{l}}_2$added.

So concentration of cation vacancies

$=10^{-4}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{\%}=\frac{10^{-4}}{100}\times 6.023\times 10^{23}\times 1=6.023\times 10^{17}$