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

Density,

$d=\frac{Z\times M}{a^3\times N_A}$

$\begin{array}{l}d=\frac{2\times 120}{64\times {10}^{-24}\times 6\times {10}^{23}}\\ \mathrm{d}=\frac{40}{6.4}=\frac{1}{0.16}\mathrm{g}/{\mathrm{cm}}^3=6.25\mathrm{g}/{\mathrm{cm}}^3\end{array}$

$\text{ Number of moles }=\frac{\text{ mass }}{\text{ molar mass }}$

Number of unit cells,

$$\begin{gathered} \mathrm{n}=\frac{12\times {10}^{22}}{2} \\ \mathrm{n}=6\times {10}^{22} \end{gathered}$$

It is known that,

$\sqrt{3}a=4r$

$r=\frac{\sqrt{3}}{4}\times 400$

$\mathrm{r}=1.732\times 100\mathrm{pm}=1.732\text{Å}$

Also,

$\text{ Volume of the unit cell }=\frac{\text{ mass }}{\text{ density }}=\frac{25}{6.25}$

$V_{\text{occupied }}=0.68\times \frac{25}{6.25}=2.72{\mathrm{cm}}^3$