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KBr is doped with $$10^{-5}$$ mole percent of SrBr$$_2$$. The number of cationic vacancies in 1 g of KBr crystal is $$10^{14}$$ ________. (Round off to the Nearest Integer). [Atomic Mass: K: 39.1u, Br: 79.9u, $$N_A = 6.023 \times 10^{23}$$]
Correct Answer: 5
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