If NaCl is doped with 10^–3 mol % of SrCl₂, what is the concentration of cation vacancies?

We know that two Na ²⁺ ions are replaced by each of the Sr²⁺ ions while SrCl₂ is doped with NaCl. But in this case only one lattice point is occupied by each other of the Sr²⁺ ion and produce one cation vacancy. Here 10^-3 mole of SrCl₂ is doped with 100 moles of NaCl.

Thus, cation vacancies produced by NaCl = 10 ^-3mol

Therefore, 1 mole of NaCl will produce cation vacancies after doping

= = 10^-5 mol

Therefore, total cationic vacancies

= 10^-5 × Avogadro’s number

= 10^{-5 ×} 6.022 × 10 ²³

= 6.023 × 10 ^-18 vacancies.