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