If `NaCl` is doped with `10^(-4)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`
A. `6.02xx10^(16) “mol”^(-1)`
B. `6.02xx10^(17) “mol”^(-1)`
C. `6.02xx10^(14) “mol”^(-1)`
D. `6.02xx10^(15) “mol”^(-1)`
A. `6.02xx10^(16) “mol”^(-1)`
B. `6.02xx10^(17) “mol”^(-1)`
C. `6.02xx10^(14) “mol”^(-1)`
D. `6.02xx10^(15) “mol”^(-1)`
Correct Answer – B