A) \[6.02\times {{10}^{6}}mo{{l}^{-1}}\]
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 :
When \[SrC{{l}_{2}}\]is doped with\[NaCl.\] One \[S{{r}^{2+}}\]replaces two \[N{{a}^{+}}\] ions and occupies a lattice point and produces one cation vacancy. 100 mol of NaCl will have \[{{10}^{-4}}\]cation vacancy \[1\,mol=\frac{{{10}^{-4}}}{100}={{10}^{-6}}\,mol\] Number of cation vacancies \[={{10}^{-6}}\times 6.02\times {{10}^{23}}\] \[=6.02\times {{10}^{17\,}}\]atoms
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