$A$ compound is formed by two elements $X$ and $Y$. Atoms of the element $Y$ (as anions) make $ccp$ and those of the element $X$ (as cations) occupy all the octahedral voids. What is the formula of the compound?

How much part of an atom occupies each corner of a $bcc$ unit cell?

Calculate the volume occupied by a particle in $bcc$ unit cell if the volume of unit cell is $8.2 \times 10^{-23} \ cm^3$.

In a close-packed structure of mixed oxides,the lattice is composed of oxide ions. One-eighth of the tetrahedral voids are occupied by divalent cations $(A^{2+})$,while one-half of the octahedral voids are occupied by trivalent cations $(B^{3+})$. The formula of the mixed oxide is:

$A$ solid has $CCP$ arrangement having atoms $A, B, C$. If $A$ atoms are present at face centres,$B$ at corners and $C$ atoms occupy $50\%$ tetrahedral voids,then the molecular formula of the solid will be:

$A$ compound forms a hexagonal close-packed $(HCP)$ structure. How many octahedral voids are present in $0.3 \, mol$ of it?