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Question
If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies?
Solution
It is given that NaCl is doped with 10−3 mol% of SrCl2.
This means that 100 mol of NaCl is doped with 10−3 mol of SrCl2.
Therefore, 1 mol of NaCl is doped with `10^(-3)/100` mol of SrCl2
= 10−5 mol of SrCl2
Cation vacancies produced by one Sr2+ ion = 1
Concentration of the cation vacancies Produced by 10-5 mol of Sr2+ ions = `10^(-5)xx6.022xx10^(23)`
=6.022 x 1018 mol-1
Hence, the concentration of cation vacancies created by SrCl2 is 6.022 × 108 per mol of NaCl.
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