Chemistry · 2021
JEE Main · 17 March 2021, Shift 2 · Q52
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…
KBr is doped with $\displaystyle 10^{-5}$ mole percent of $\displaystyle \mathrm{SrBr}_2$. The number of cationic vacancies in $\displaystyle 1$ g of KBr crystal is \_\_\_\_ $\displaystyle 10^{14}$. (Round off to the Nearest Integer).
[Atomic Mass : K : $\displaystyle 39.1$ u, Br : $\displaystyle 79.9$ u
$$\left.\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}\right]
$$
Official answer
5
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.