If NaCl is doped with 10–3 mole of BaCl2. What is the concentration of cationic vacancies?

NaCl is doped with 10-3 mole % of SrCl2.This means, 100 moles of NaCl are doped with 10-3 moles of SrCl2.

Therefore, 1mole of NaCl is doped with 10-3 /100 = 10-5 moles of SrCl2 or 10-5 moles of Sr2+.

Now,

One Sr2+ ion create one cation vacancy.

Therefore, no. of cation vacancies created by 10-5 moles Sr2+

= 10-5 moles/mole of NaCl

= 10-5 x 6.022 x 1023 / mole of NaCl

= 6.022 x 1018/ mole of NaCl