Calculate the density of an element having molar mass $27 \ g \ mol^{-1}$ that forms $fcc$ unit cell. $[a^3 \cdot N_A = 38.5 \ cm^3 \ mol^{-1}]$ (in $g \ cm^{-3}$)

$A$ metal crystallizes with a $FCC$ lattice, the edge of whose unit cell is $x \text{ pm}$. The diameter of this metal atom would be $\text{pm}$.

$A$ certain element crystallises in a $bcc$ lattice of unit cell edge length $27 \mathring{A}$. If the same element under the same conditions crystallises in the $fcc$ lattice,the edge length of the unit cell in $\mathring{A}$ will be .........
(Round off to the Nearest Integer).
[Assume each lattice point has a single atom]
[Assume $\sqrt{3}=1.73, \sqrt{2}=1.41$]

Calculate the number of atoms present per unit cell if the product of density and volume of the unit cell is $1.8 \times 10^{-22} \ g$. [Mass of an atom $= 4.5 \times 10^{-23} \ g$]

Sodium metal crystallizes in $B.C.C.$ lattice with an edge length of $4.29 \ \mathring{A}$. The radius of the sodium atom is:

Calculate the volume of a unit cell having four particles in it with a density of $19.0 \ g \ cm^{-3}$ [molar mass of element $= 190 \ g \ mol^{-1}$].