A) \[6.02\times {{10}^{15}}mo{{l}^{-1}}\]
B) \[6.02\times {{10}^{16}}mo{{l}^{-1}}\]
C) \[6.02\times {{10}^{17}}mo{{l}^{-1}}\]
D) \[6.02\times {{10}^{14}}mo{{l}^{-1}}\]
Correct Answer: A
Solution :
| [a] If \[NaCl\,(N{{a}^{+}})\] is doped with \[{{10}^{-4}}\] mol % of \[SrC{{l}_{2}}\,(S{{r}^{2+}})\] |
| \[2N{{a}^{+}}\] ion doped by \[S{{r}^{2+}}\,{{N}_{A}}=6.02\times {{10}^{23}}\] |
| The concentration of cation vacancies = |
| \[=6.02\times {{10}^{23}}\times {{10}^{-8}}\] |
| \[=6.02\times {{10}^{15}}\,mo{{l}^{-1}}\] |
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