If NaCl is doped with 10^-3 mol% SrCl2, then the concentration of cation vacancies will be
a) 1 x 10^-3 mol%
b) 2 x 10^-3 mol%
c) 3 x 10^-3 mol%
d) 4 x 10^-3 mol%
URGENT PLEASE!!!
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No. of cationic vacancies will be eq to no of Sr2+ ions which have replaced 2 Na+ ions. So, conc. of cationic vacancies = conc. of SrCl2 = 10^-3mol%(a).
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