Question

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

Answer 1

When NaCl is doped with SrCl₂, each additional Sr²⁺ ion eliminates two Na⁺ ions, creating two vacancies, one occupied by Sr²⁺ and the other unoccupied. Thus, adding one Sr²⁺ ion results in one cation vacancy.

Doping of NaCl with 1 × 10⁻³ rnol % of SrCl₂ means that 1 × 10⁻³ moles of NaCl are added to 100 moles of NaCl. 

∴ 1 mole of NaCl contains SrCl₂ = `(1 xx 10^-3)/100`

= 1 × 10⁻⁵ moles

= 1 × 10⁻⁵ × 6.022 × 10²³ molecules

= 6.022 × 10¹⁸ molecules

∴ No. of Sr²⁺ ions present in one mole of NaCl = No. of cation vacancies in one mole of NaCl

= 6.022 × 10¹⁸