In a crystal lattice,the number of octahedral voids compared to tetrahedral voids is ...............

Atoms of element $A$ occupy $\frac{1}{3}$ of the octahedral voids,while atoms of element $B$ form an $hcp$ structure. Determine the formula of the compound formed by $A$ and $B$.

In a compound $AB_2O_4$,the oxide ions form a cubic close-packed $(CCP)$ lattice. The molar ratio of trivalent cations to bivalent cations is $2:1$. If all the bivalent ions occupy octahedral voids and trivalent ions occupy both tetrahedral and octahedral voids in equal numbers,then the fraction of octahedral voids that remain unoccupied is:

The volume occupied by the atoms in a unit cell of hexagonal close packing $(HCP)$ or cubic close packing $(CCP)$ is equal to:

The volume of the $HCP$ unit cell is

$A$ compound of formula $A_2B_3$ has the $hcp$ lattice. Which atom forms the $hcp$ lattice and what fraction of tetrahedral voids is occupied by the other atoms?