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 mol of NaCl is doped with \(\frac{10^{-3}}{100}\) mol of SrCl₂

= 10^-3 mol% of SrCl₂

Cation vacancies produced by one Sr²⁺ ion = 1

∴ Concentration of the cation vacancies produced by 10^-5 mol of Sr²⁺ ion = 10^-5 x 6.022 x 10²³

= 6.022 x 10¹⁸mol^-1

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