Question

If NaCl is doped with ${10}^{-4}$ mole % of $SrC{l}_2$, the concentration of cation vacancies will be :

• A. $6.02\times {10}^{12}mo{l}^{-1}$
• B. $6.02\times {10}^{17}mo{l}^{-1}$
• C. $6.02\times {10}^{14}mo{l}^{-1}$
• D. $6.02\times {10}^{15}mo{l}^{-1}$

Answer 1

The correct option is B $6.02\times {10}^{17}mo{l}^{-1}$

Dopping if $SrC{l}_2$ to NaCl brings in replacement of two $N{a}^{+}$ by each $S{r}^{2+}$ ion, but $S{r}^{2+}$ occupies one lattice point. This produces one cation vacancy.
No. of cation vacancies = ${10}^{-4}$
100 mole of NaCl will have cationic vacancy = ${10}^{-4}$
$\therefore$ 1 mole of NaCl will have cationic vacancy
$\therefore \frac{{10}^{-4}}{100}={10}^{-6}$
$\therefore$ No. of cationic vacancies = ${10}^{-6}\times 6.02\times {10}^{23}=6.02\times {10}^{17}$