If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.023 x 1023 mol-1)
(a) 6.023 x 1015 mol-1 (b) 6.023 x 1016 mol-1
(c) 6.023 x 1017 mol-1 (d) 6.023 x 1014 mol-1