If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?

If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration

1.	If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?
Answer» Since for addition of one Sr²⁺ ion two Na⁺ ion need to be displaced i.e. one vacancy will be created for each Sr²⁺ ion. Concentration of cation vacancies = Concentration of Sr²⁺ ion = 6.022 × 10²³ × \(\frac { { 10 }^{ -3 } }{ 100 } \) ions/mole of NaCl = 6.022 × 10¹⁸ ions/mole of NaCl Hence concentration of cation vacancies is 6.022 × 10¹⁸ vacancies/mole of NaCl

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If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?

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