Question

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

Answer 1

$N{a}^{+}$	$C{l}^{-}$	$N{a}^{+}$	$C{l}^{-}$
$C{l}^{-}$		$C{l}^{-}$	$N{a}^{+}$
$S{r}^{2+}$	$C{l}^{-}$	$N{a}^{+}$	$C{l}^{-}$
$C{l}^{-}$	$N{a}^{+}$	$C{l}^{-}$	$N{a}^{+}$

Number of cationic vacancies $=\frac{{10}^{-3}\times 6.023\times {10}^{23}}{100}=6.023\times {10}^{18}$ vacancies per mol