An element has a cubic close packing (ccp) structure with an edge length of \(297\, \ pm\). Calculate the density of the element. (Given: Molar mass is \(107.9\, \ gmol^{-1}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2.7 × 10^4\ kg\,m^{-3}\)
The density of a crystal equals the density of its unit cell:
\( d = \displaystyle\frac{z\,M}{a^3\,N_A} \)
where z is the number of atoms per unit cell, M the molar mass, a the edge length and NA the Avogadro constant. A ccp (fcc) lattice has z = 4.
a = 297 pm = 2.97 × 10−8 cm
a3 = (2.97 × 10−8)3 = 2.62 × 10−23 cm3
d = (4 × 107.9) / (2.62 × 10−23 × 6.022 × 1023) g cm−3
d = 431.6 / 15.78 = 27.4 g cm−3
d = 27.4 × 103 kg m−3 ≈ 2.7 × 104 kg m−3
(1 g cm−3 = 103 kg m−3.)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.