JEE Main 2020ChemistrySolid StateMediumNumerical

JEE Main 2020 — Solid State Question with Solution

JEE Main 2020 (03 Sep Shift 1)

Question

An element with molar mass 2.7×10−2 kg mol−1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7×103 kg m-3, the radius of the element is approximately ……….×10-12m (to the nearest integer)

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Show full solutionCorrect answer: 143
Correct answer
143

Step-by-step explanation

Density = Z×MNa×Volume

d=Z×276.02×1023×4.05×10-33

2.7=Z×276.02×1023×4.05×10-33

Z=4  ⇒fcc unit cell

For fcc 

unit cell  4r=2a

r=1.414×4054

=143.1675 pm

=143.17 pm

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About this question

This is a previous-year question from JEE Main 2020, covering the Solid State chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.