An element A (Atomic weight = 120) having bcc structure has unit edge length 400 pm. Identify the correct option(s). (Given: $N_A=6\times {10}^{23}$)

The correct option is D In 25 g of solid element, the volume occupied by atoms is nearly $2.72c{m}^3$ only
(A) Density of the solid element
$d=\frac{Z\times M}{a^3\times N_A}$
Where,
$Z$ is the number of atoms per unit cell
$M$ is molar mass of the atom
$a$ is the edge length of unit cell
$N_A$ is the Avogadro number
Given,
$Z$ for BCC structure = 2
Molar mass = 120 g/mol
Edge length = $400\times {10}^{-10}$ cm

$d=\frac{2\times 120}{64\times {10}^{-24}\times 6\times {10}^{23}}$

$=\frac{40}{6.4}=\frac{1}{0.16}g/c{m}^3=6.25g/c{m}^3$

(B) Number of atoms per unit cell for BCC = 2
Number of unit cell $=\frac{\text{number of atoms}}{2}$
Number of atoms $=\text{number of moles}\times N_A$

Number of moles $=\frac{\text{mass}}{\text{molar mass}}$
Given mass = 24 g
number of moles $=\frac{24}{120}=0.2$
Number of atoms $=0.2\times 6\times {10}^{23}$

Number of unit cell $=\frac{12\times {10}^{22}}{2}$
$=6\times {10}^{22}$

(C) The relationship between radius and edge length for BCC is given by
$\sqrt{3}a=4r\Rightarrow r=\frac{\sqrt{3}}{4}\times 400$
$=1.732\times 100pm=1.732\stackrel{\circ }{A}$

(D) Given mass = 25 g and density = $6.25g/c{m}^3$
Volume of the unit cell $=\frac{\text{mass}}{\text{density}}=\frac{25}{6.25}$
Fraction of the volume occupied by atoms in BCC = 0.68
hence,
$V_{\text{occupied}}=0.68\times \frac{25}{6.25}=2.72c{m}^3$