Question

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

• A. $6.02\times {10}^{20}$
• B. $6.02\times {10}^{22}$
• C. $6.02\times {10}^{23}$
• D. $6.02\times {10}^{18}$

Answer 1

The correct option is D $6.02\times {10}^{18}$

The difference in charge on $S{r}^{2+}$ and on $N{a}^{+}$ is $1$. Substitution of $1N{a}^{+}$ ion with $1S{r}^{2+}$ ion causes one cationic vacancy.
Doping of NaCl with ${10}^{-3}$ mole % $S{r}^{2+}$ creates vacancies
$=6.023\times {10}^{23}/mol$ $\times \frac{{10}^{-3}}{100}{mol}^{-1}$
$=6.023\times {10}^{18}$cation -vacancies.
Hence, (d) is correct.