Solid State Question 267
Question: If NaCl is doped with $ {{10}^{-3}} $ mole% of $ SrCl_2 $ , the number of cationic vacancies is
Options:
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}} $
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Answer:
Correct Answer: D
Solution:
[d] Due to the addition of $ SrCl_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. $ SrCl_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}} $
BETA
