JEE Main 2021 — Solid State Question with Solution
From: JEE Main 2021 (Online) 17th March Evening Shift
Question
KBr is doped with 105 mole percent of SrBr2. The number of cationic vacancies in 1g of KBr crystal is ____________ 1014. (Round off to the Nearest Integer).
[Atomic Mass : K : 39.1 u, Br : 79.9 u NA = 6.023 1023]
[Atomic Mass : K : 39.1 u, Br : 79.9 u NA = 6.023 1023]
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Show full solutionCorrect answer: 5
Correct answer
5
Step-by-step explanation
For every Sr+2 ion, 1 cationic vacancy is created. Hence, no. of Sr+2 ion = Number of cationic vacancies
Since mole percentage of SrBr2 dropped is 105 to that of total moles of KBr.
Hence,
No. of cationic vacancy
Since mole percentage of SrBr2 dropped is 105 to that of total moles of KBr.
Hence,
No. of cationic vacancy
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