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If NaCl is doped with 10−4 mol % of SrCl2, the concentration of cation vacancies will be (NA=6.02×1023 mol−1)

1. 6.02×1016 mol−1
2. 6.02×1017 mol−1
3. 6.02×1014 mol−1
4. 6.02×1015 mol−1

Correct answer: 6.02×1017 mol−1

Solution

When SrCl2 is doped into NaCl, Sr2+ replaces Na+ ions, creating one cation vacancy for every Sr2+ ion introduced. For 10^-4 mol% doping, the number of Sr2+ ions is (10^-4/100) × 1 mol × NA = 6.02 × 10^17 vacancies per mole.

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