1.

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

Answer»

NaCl is doped with 10-3 mol % of SrCl2
∴ In 100 part of NaCl = 10-3 mol SrCl2
1 part of NaCI = \(\frac { 10^{ -3 } }{ 100 }\) mol Srcl2
= 10-5 mol Srcl2
= 6.022 x 1023 x 10-5 Srcl2
= 6.022 x 1018 Srcl2
As each Sr2+ ion creates one void hence number of voids
= 6.022 x 1018



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