Given, concentration of SrCl2=10−3 mol\% concentration is in percentage so that take total 100‌mL of solution.Number of moles of NaCl=100−3 moles of SrCl2Moles of SrCl2 is very negligible as compare to total moles so percentage always taken as 100 so that 1 mole of NaCl is dipped with =
10−3
100
mole of SrCl2 =10−5‌mole‌of‌SrCl2 So, cation vacancies per mole of NaCl=10−5‌mol Since, 1‌mole=6.022×1023 particles So, cation vacancies per mole of NaCl =10−5×6.022×1023 =6.022×1018