Question

If NaCl is doped with 10⁻² mol percentage of strontium chloride, what is the concentration of cation vacancy?

Answer 1

We know that two Na+ ions are replaced by each of the Sr²⁺ ions while SrCl₂, is doped with NaCI. But in this case, only one lattice point is occupied by each of the Sr²⁺ ions and produces one cation vacancy.

Here 10⁻² mole of SrCl₂, is doped with 100 moles of NaCI. Thus, cation vacancies produced by NaCi = 10⁻²  mol. Since 100 moles of NaCl produces cation vacancies after doping = 10⁻² mol.

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

= `10^-2/100` = 10⁻⁴ mol

∴ Total cationic vacancies,

= 10⁻⁴ × Avogadro’s number

= 10 × 6.023 × 10²³

= 6.023 × 10¹⁹ vacancies