Question

If NaCl is doped with 10⁻⁴ mol % of SrCl₂, the
concentration of cation vacancies will be (NA = 6.02 × 10²³ mol⁻¹)
(a) 6.02 × 10¹⁶ mol⁻¹ (b) 6.02 × 10¹⁷ mol⁻¹
(c) 6.02 × 10¹⁴ mol⁻¹ (d) 6.02 × 10¹⁵ mol⁻¹

Answer 1

(d) 6.02 × 10¹⁵ mol⁻¹

Explanation:

If NaCl is doped with 10⁻⁴ mol % of SrCl₂, the concentration of cation vacancies will be:

Sr2+ = 2 * 10^-4 = 10^-8

Na = 6.02 × 10²³

So concentration of cation vacancies =  6.02 × 10²³ * 10^-8 =  6.02 × 10¹⁵ mol⁻¹

Option D is the answer.