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Question 25

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


1 Cation of Sr2+ will create 1 cation vacancy in Nacl.Therefore the number of cation vacancies created in the lattice of Nacl is equal to the number of divalent Sr2+ ions added.

Now the concentration of cation vacancy on being doped with 10-3 mol% of Srcl2.

= 10-3 mol% = 10-3 / 100 = 10-5 mol\

Also number of Sr2+ ion in 10-5 mol = 10-5 x 6.023 x 1023 = 6.023 x 1018

Therefore number of cation vacancies = 6.023 x 1018

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