Question

If NaCl is doped with 10^–3 mol % of SrCl₂ calculate the concentration of cation vacancies.

Options

• 5.023 × 10¹⁷

• 6.022 × 10²³

• 6.022 × 10¹⁸

• 5.022 × 10¹⁸

Answer 1

6.022 × 10¹⁸ 

Explanation:

Due to the addition of strontium chloride (SrCl₂) each Sr²⁺ ion replaces two Na⁺ ions but occupies only one lattice point in place of Na⁺ ion. This leads to one cation vacancies.

No of moles of cation vacancies is 100 mol of Nacl = 10^–3

No of moles of cation vacancies is 1 mol = `10^-3/100` = 10^–5 mol

Total no. of cation vacancies = 10^–5 × N_A

= `(10^-5  "mol") xx (6.022 xx 10^23  "mol"^-1)`

= 6.022 × 10¹⁸