Solid State Question 40
Question: If $ NaCl $ is doped with $ {{10}^{-4}}mol% $ of $ SrCl_2 $ , the concentration of cation vacancies will be $ (N _{A}=6.02\times 10^{23}mo{{l}^{-1}}) $
Options:
A) $ 6.02\times 10^{16}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}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Since each $ S{{r}^{++}} $ ion provides one cation vacancy, hence Concentration of cation vacancies = mole % of $ SrCl_2 $ added $ ={{10}^{-4}}mole $ % $ =\frac{{{10}^{-4}}}{100}\times 6.023\times 10^{23}=6.023\times 10^{17} $
BETA
