Question

$NaCl$ is doped with ${10}^{-6}mol$ % $Sr{Cl}_2$, the concentration of cation vacancies is:
[Heat concentration of vacancies $=\frac{{10}^{-4}}{100}{N}_A$]

• A. $6.02\times {10}^{15}$ ${mol}^{-1}$
• B. $6.02\times {10}^{16}$ ${mol}^{-1}$
• C. $6.02\times {10}^{17}$ ${mol}^{-1}$
• D. $6.02\times {10}^{14}$ ${mol}^{-1}$

Answer 1

The correct option is A $6.02\times {10}^{15}$ ${mol}^{-1}$

Total $SrC{l}_2$ used= ${10}^{-6}\times {10}^{-2}={10}^{-8}mol$

Moles of cations= ${10}^{-8}mol$

$1$ cation $\left(S{r}^{2+}\right)$ creates $1$ cation vacancy

$\therefore$ Moles of cation vacancy= ${10}^{-8}mol$

Concentration of cation= $N_A\times$ Number of moles

$=6.02\times {10}^{23}\times {10}^{-8}$

$=6.02\times {10}^{15}$