Q. If $NaCl\$is doped with $10^{- 3}$mol % $SrCl_{2}$, then the concentration of cation vacancies will be
A) $1 \times 10^{- 3}$mol %
B) $2 \times 10^{- 3}$mol %
C) $3 \times 10^{- 3}$ mol %
D) $4 \times 10^{- 3}$ mol %
Explanation
(a)
At each $Sr^{2 +}$ ion introduces one cation vacancy, therefore, concentration of cation vacancies = mol % of$SrCl_{2}$ added
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.