If NaCl is doped with

It is given that NaCl is doped with 10−3 mol% of SrCl2.

This means that 100 mol of NaCl is doped with 10−3 mol of SrCl2.

Therefore, $1 \mathrm{~mol}$ of $\mathrm{NaCl}$ is doped with $\frac{10^{-3}}{100} \mathrm{~mol}$ of $\mathrm{SrCl}_{2}$

= 10−5 mol of SrCl2

Cation vacancies produced by one Sr2+ ion = 1

$\therefore$ Concentration of the cation vacancies

produced by $10^{-5} \mathrm{~mol}$ of $\mathrm{Sr}^{2+}$ ions $=10^{-5} \times 6.022 \times 10^{23}$

$=6.022 \times 10^{18} \mathrm{~mol}^{-1}$

Hence, the concentration of cation vacancies created by SrCl2 is 6.022 × 108 per mol of NaCl.