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| 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. `6.02xx10^(15)"mol"^(-1)`B. `6.02xx10^(16)"mol"^(-1)`C. `6.02xx10^(17)"mol"^(-1)`D. `6.02xx10^(14)"mol"^(-1)` |
| Answer» Correct Answer - C | |