If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.023 x 1023 mol-1)
(c) Doping of NaCl with 10-4 mol % of SrCl2 means, 100 moles of NaCl are doped with 10-4 mol of SrCl2.
... 1 mol of NaCl is doped with
SrCl2 = 10-4/100 = 10-6 mole
As each Sr2+ ion introduces one cation vacancy.
... Concentration of cation vacancies
= 10-6 mol/mol of NaCl
= 10-6 x 6.023 x 1023 mol-1
= 6.023 x 1017 mol-1