Question

IF $\mathrm{NaCl}$ is doped with $10^{-4}$ mol $\%$ of $\mathrm{SrCl}_{2}$, the concentration of cation vacancies will $b e\left(\mathrm{~N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)$
(a) $6.02 \times 10^{16} \mathrm{~mol}^{-1}$
(b) $6.02 \times 10^{17} \mathrm{~mol}^{-1}$
(c) $6.02 \times 10^{14} \mathrm{~mol}^{-1}$
(d) $6.02 \times 10^{15} \mathrm{~mol}^{-1}$

Answer 1

Correct option (b) $6.02 \times 10^{17} \mathrm{~mol}^{-1}$
Explanation:
As each $\mathrm{Sr}^{2+}$ ion introduces one cation vacancy, therefore, concentration of cation vacancies $=$ mole $\%$ of $\mathrm{SrCl}_{2}$, added.
$\therefore$ Concentration of cation vacancies
$$\begin{aligned} &=10^{-4} \mathrm{~mole} \% \\ &=\frac{10^{-4}}{100} \times 6.023 \times 10^{23} \\ &=6.023 \times 10^{17} \end{aligned}$$