A) \[6.02\times {{10}^{18}}\text{ }mo{{l}^{-1}}\]
B) \[6.02\times {{10}^{15}}\text{ }mo{{l}^{-1}}\]
C) \[6.02\times {{10}^{21}}\text{ }mo{{l}^{-1}}\]
D) \[6.02\times {{10}^{12}}\text{ }mo{{l}^{-1}}\]
Correct Answer: A
Solution :
Each \[S{{r}^{2+}}\] ion introduces one cationic vacancy, therefore, number of cationic vacancies = \[{{10}^{-3}}\] mole % \[=\frac{{{10}^{-3}}}{100}\times 6.02\times {{10}^{23}}\,mo{{l}^{-1}}\] \[=\text{ }6.02\times {{10}^{18}}\text{ }mo{{l}^{-1}}\]You need to login to perform this action.
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