The distance between two parallel $(111)$ planes in a crystal with an edge length of $300 \, pm$ is ................ $pm$.

$A$ diatomic molecule $X_2$ has a body-centred cubic (bcc) structure with a cell edge of $300 \ pm$. The density of the molecule is $6.17 \ g \ cm^{-3}$. The number of molecules present in $200 \ g$ of $X_2$ is (Avogadro constant $N_A = 6 \times 10^{23} \ mol^{-1}$) (in $N_A$)

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the number of atoms present in one unit cell of it is (Given : $N_{A}=6.023 \times 10^{23} \ mol^{-1}$)

Sodium metal crystallizes as a body-centered cubic $(bcc)$ lattice with the cell edge $4.29 \ \mathring{A}$. What is the radius of a sodium atom?

The density of $KBr$ is $2.75 \, g \, cm^{-3}$ and the unit cell edge length is $654 \, pm$. Given the atomic masses $K = 39$ and $Br = 80$, what is the crystal structure of $KBr$?

Calculate the number of particles present per unit cell if the mass of a particle is $8.0 \times 10^{-23} \ g$ and the product of density and volume of the unit cell $(\varrho \times a^3)$ is $3.2 \times 10^{-22} \ g$.