A) \[6.02\times {{10}^{16}}mo{{l}^{-1}}\]
B) \[6.02\times {{10}^{17}}mo{{l}^{-1}}\]
C) \[6.02\times {{10}^{14}}mo{{l}^{-1}}\]
D) \[6.02\times {{10}^{15}}mo{{l}^{-1}}\]
Correct Answer: B
Solution :
[b] Since each \[S{{r}^{++}}\] ion provides one cation vacancy, hence Concentration of cation vacancies = mole % of \[SrC{{l}_{2}}\] added \[={{10}^{-4}}\text{mole}\]% \[=\frac{{{10}^{-4}}}{100}\times 6.023\times {{10}^{23}}=6.023\times {{10}^{17}}\]You need to login to perform this action.
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