Solid State Question 64
Question: If NaCl is doped with mol% $ SrCl_2, $ what is the concentration of cation vacancies-
Options:
A) $ 6.02\times 10^{18}mo{{l}^{-1}} $
B) $ 6.02\times 10^{19}mo{{l}^{-1}} $
C) $ 6.02\times 10^{23}mo{{l}^{-1}} $
D) $ 6.02\times 10^{26}mo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Doping of NaCl with $ {{10}^{-3}} $ mol% $ SrCl_2 $ means that 100 moles of NaCl are doped with $ {{10}^{-3}} $ mol of $ SrCl_2 $ .
$ \therefore $ 1 mole of NaCl is doped with $ SrCl_2=({{10}^{-3}}/100) $ mol $ ={{10}^{-5}} $ mol
As each $ S{{r}^{2+}} $ ion introduces one cation vacancy, therefore concentration of cation vacancies $ ={{10}^{-5}}\text{mol/mol} $ of NaCl $ ={{10}^{-5}}\times 6.02\times 10^{23}mo{{l}^{-1}} $ $ =6.02\times 10^{18}mo{{l}^{-1}} $
BETA
