1.

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. `6.02 xx 10^(14) mol^(-1)`B. `6.02 xx 10^(15) mol^(-1)`C. `6.02 xx 10^(16) mol^(-1)`D. `6.02 xx 10^(17) mol^(-1)`

Answer» Correct Answer - D


Discussion

No Comment Found

Related InterviewSolutions