JEE Main 2021 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 17 March 2021, Shift 2, Chemistry Q52

Question 52 of 90 in this shift, Chemistry question 22 of 30, Section B.

KBr is doped with

10^{-5}

mole percent of

\mathrm{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 : 39.1 u, Br : 79.9 u

\mathrm{N_A} = 6.023 \times 10^{23}

]

Official answer

NTA final key.

Chapter: The Solid State
Topic: Solid State Defects
Idea tested: Impurity Defects and Doping

Same topic in other shifts

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.