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 :
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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