• question_answer If NaCl is dropped with ${{10}^{-4}}$ mole % of $SrC{{l}_{2}},$the concentration of cation vacancies will be 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) $6.02\times {{10}^{15}}mo{{l}^{-1}}$

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