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JEE-MAIN 2021
17-03-2021 S2
Question
KBr is doped with $10^{-5}$ mole percent of $\mathrm{SrBr}_2$. The number of cationic vacancies in 1 g of KBr crystal is $\_\_\_\_$ 10¹4. (Round off to the Nearest Integer).
[Atomic Mass : K : $39.1 \mathrm{u}, \mathrm{Br}: 79.9 \mathrm{u}$, $$ \left.N_A=6.023 \times 10^{23}\right] $$
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JEE Main
Physics
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