Question

If NaCl is doped with 10^–4 mole percent of SrCl₂ the concentration of cation vacancies will be

पर्याय

• 6.022 × 10^–16 mo1^–1

• 6.022 × 10¹⁷ mo1^–1

• 6.022 × 10¹⁴ mo1^–1

• 6.022 × 10¹³ mo1^–1

Answer 1

6.022 × 10¹⁷ mo1^–1 

Explanation:

Doping by each Sr²⁺ ion will create one cation vacancy.

∴ Concentration of cation vacancies = 6.022 × 10²³ × = 6.022 x 10¹⁷.