If NaCl is doped with 10⁻³ mol % of SrCl₂ what is the concentration of cation vacancies?
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NaCl is doped with 10 −3 mol% of SrCl 2 so 100 mol of NaCl is doped with 10 −3 mol of SrCl 2;
=> 1 mol of NaCl is doped with 10^-3/1000 mol of SrCl2; =10^-5;
No of Cation vacancies produced by one Sr 2+ ion = 1;
no of cation vacancies produced by 10^-5 mole of SrCl2 =10^-5*6.022*10^23; =6.022*10^18 per mole
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