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JEE Main 20203 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
3 September
Shift
Morning Shift
Subject
Chemistry
Chapter
Solid State

JEE Main 2020 · 3 September · Morning Shift · Chemistry · Solid State

An element with molar mass 2 .7 × 10 ^ - 2 kg mol ^ - 1 forms a...

An element with molar mass 2.7×10−2kgmol−12 .7 \times {10}^{- 2}kg{mol}^{- 1} forms a cubic unit cell with edge length 405pm.405pm . If its density is 2.7×103kgm−32 . 7 \times {10}^{3}kg{m}^{- 3}, the radius of the element is approximately ……….×10−12m\ldots \ldots \ldots . \times {10}^{- 12}m (to the nearest integer)

Correct answer

0

Step-by-step solution

Density = Z×MNa×Volume\dfrac{Z \times M}{{N}_{a} \times Volume}

d=Z×276.02×1023×[4.05×10−3]3d = \dfrac{Z \times 27}{6 . 02 \times {10}^{23} \times {\left[4 . 05 \times {10}^{- 3}\right]}^{3}}

2.7=Z×276.02×1023×[4.05×10−3]32 . 7 = \dfrac{Z \times 27}{6 . 02 \times {10}^{23} \times {\left[4 . 05 \times {10}^{- 3}\right]}^{3}}

Z=4⇒fccZ =4 ⇒ fcc unit cell

For fccfcc

unit cell 4r=2a4r = \sqrt{2}a

r=1.414×4054r = \dfrac{1 . 414 \times 405}{4}

=143.1675pm= 143 . 1675pm

=143.17pm= 143 . 17pm

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