A) \[6.02\times {{10}^{-18}}mo{{l}^{-1}}\]
B) \[{{10}^{-5}}mo{{l}^{-1}}\]
C) \[6.02\times {{10}^{20}}mo{{l}^{-1}}\]
D) \[6.02\times {{10}^{18}}mo{{l}^{-1}}\]
Correct Answer: D
Solution :
[d] Due to the addition of\[SrC{{l}_{2}}\], each \[S{{r}^{2+}}\]ion replaces two \[N{{A}^{+}}\]ions, but occupies one \[N{{A}^{+}}\]lattice point. Thus, this exchange of \[N{{a}^{+}}\]ion by \[S{{r}^{2+}}\]ion makes one cationic vacancy. \[SrC{{l}_{2}}\]doped\[={{10}^{-3}}\]mol per 100 mol \[={{10}^{-5}}\]mol per 1 mol \[\therefore \]Cation vacancies \[={{10}^{-5}}\]mol per 1 mol \[={{10}^{-5}}\times {{N}_{A}}mo{{l}^{-1}}={{10}^{-5}}\times 6.02\times {{10}^{23}}\] Total \[=6.02\times {{10}^{18}}\]cationic vacancies \[mo{{l}^{-1}}\]
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