JEE Main 17 March 2021 Shift 2 Solved Paper

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