The volume of a simple cubic unit cell is $x \times 10^{-23} \ cm^3$. Calculate the value of $x$ if the volume occupied by a particle in it is $2.1 \times 10^{-23} \ cm^3$.

$A$ metal crystallises with a face-centred cubic $(fcc)$ lattice. The edge length of the unit cell is $408 \, pm$. The diameter of the metal atom is ............. $pm$.

Elements $A$ and $B$ have $fcc$ and $bcc$ structures respectively with a unit cell edge length of $3 \mathring{A}$ for both elements. The number of atoms in $210 \ g$ of $A$ is equal to the number of atoms in $594 \ g$ of $B$. If the density of $A$ is $7 \ g \ cm^{-3}$,what is the density of $B$ (in $g \ cm^{-3}$)?

The edge length of a cube is $300 \ pm$. Its body diagonal would be: (in $pm$)

An element has a body-centered cubic structure with a unit cell edge length of $400 \ pm$. The atomic mass of the element is $24 \ g \ mol^{-1}$. What is the density of the element (in $g \ cm^{-3}$)? $(N_{A} = 6 \times 10^{23} \ mol^{-1})$

$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is ................. $g \, mol^{-1}$.
($N_A$ Avogadro's constant $= 6.02 \times 10^{23} \, mol^{-1}$)