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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)

6.02 × 1016 mol–1
6.02 × 1017 mol–1
6.02 × 1014 mol–1
6.02 × 1015 mol–1
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