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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. `6.02 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 | |