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
$$
\begin{aligned}
&=10^{-4} \mathrm{~mole} \% \\
&=\frac{10^{-4}}{100} \times 6.023 \times 10^{23} \\
&=6.023 \times 10^{17}
\end{aligned}
$$