Question
If NaCl is doped with 10-4 mol% of SrCl2 the concentration of cation vacancies will be
(NA = 6.02 x 1023 mol-1):
-
6.02 x 1015 mol-1
-
6.02 x 1016 mol-1
-
6.02 x 1017
-
6.02 x 1014
Solution
A.
6.02 x 1015 mol-1
If NaCl is doped with 10-4 mol% of SrCl2,
2 Na+ ions doped by Sr2+ NA = 6.02 x 1023
The concentration of cation vacancies
= 6.02 x 1023 x10-8
= 6.02 x 1015 mol-1