The number of cationic vacancies in 1g of KBr crystal is 5×1014. For every Sr2+ ion, 1 cationic vacancy is created. Hence, number of Sr2+ ion = number of cationic vacancies Since, the mole percentage of SrBr2 dopped is 10−5 to that of total moles of KBr. Hence, number of cationic vacancy =‌
10−5
100
×‌
1
119
×NA =‌
1
119
×10−7×6.022×1023 (Mass of KBr =119) =5×10−2×10−7×1023=5×1014