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Question

If NaCl is doped with 104 mole % of SrCl2, the concentration of cation vacancies will be :

A
6.02×1012mol1
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B
6.02×1017mol1
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C
6.02×1014mol1
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D
6.02×1015mol1
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Solution

The correct option is B 6.02×1017mol1
Dopping if SrCl2 to NaCl brings in replacement of two Na+ by each Sr2+ ion, but Sr2+ occupies one lattice point. This produces one cation vacancy.
No. of cation vacancies = 104
100 mole of NaCl will have cationic vacancy = 104
1 mole of NaCl will have cationic vacancy
104100=106
No. of cationic vacancies = 106×6.02×1023=6.02×1017

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