Q. 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
Explanation
[Solution from source] Each Sr²⁺ replaces two Na⁺ and creates one cation vacancy. 10⁻⁴ mol% = 10⁻⁶ mol SrCl₂ per mol NaCl, giving 10⁻⁶ × 6.02 × 10²³ = 6.02 × 10¹⁷ vacancies per mol. 6.02 × 10¹⁹ would ignore the percent conversion. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.