Question

If $NaCl$ is doped with ${10}^{-3}$ $mol$ percent of $SrC{l}_2$, what is the concentration of cation vacancies?

Answer 1

$NaCl$ is doped with ${10}^{-3}$ mol % of $SrC{l}_2$.

$100$ moles of $NaCl$ are doped with $0.001$ moles of $SrC{l}_2$.

$1$ mole of $NaCl$ is doped with $\frac{0.001}{100}={10}^{-5}$ moles of $SrC{l}_2$ or ${10}^5$ moles of $S{r}^{2+}$ cations.
One $S{r}^{2+}$ ion creates one cation vacancy.

${10}^{-5}$ moles of $S{r}^{2+}$ will create ${10}^{-5}$ moles of cation vacancies which correspond to ${10}^{-5}\times 6.023\times {10}^{23}=6.023\times {10}^{18}$ cation vacancies per mole of $NaCl$.