$Xe$ crystallizes in an $FCC$ structure. The edge length of its unit cell is $620 \, pm$. The radius of $Xe$ is $=$ ...... $pm$.

How many unit cells are present in $100 \ g$ of an element with $fcc$ crystal structure having density $10 \ g/cm^{3}$ and edge length $100 \ pm$?

Calculate the number of unit cells in $1 \ cm^3$ volume of metal if the unit cell edge length is $1.25 \times 10^{-8} \ cm$.

Ionic radii of cation $A^{+}$ and anion $B^{-}$ are $102 \ pm$ and $181 \ pm$ respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for $B^{-}$ and $A^{+}$ is present in all octahedral voids. The edge length of the unit cell of the crystal $AB$ is $pm$.

$A$ 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 constant $= 6.02 \times 10^{23} \, mol^{-1}$ )

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 $= 4.5 \times 10^{-23} \ g$]