An $fcc$ unit cell of aluminium contains the equivalent of how many atoms?

If the number of atoms per unit cell in a crystal is $2$,the structure of the crystal is:

An element crystallises in a face-centred cubic (fcc) unit cell with cell edge $a$. The distance between the centres of two nearest octahedral voids in the crystal lattice is

In a face-centered cubic $(FCC)$ arrangement,atoms $A$ are present at the corners and atoms $B$ are present at the face centers. If $4$ atoms of $A$ are removed from the corners,what is the empirical formula of the compound?

In a compound,$W$ atoms are at the corners of the unit cell,$O$ atoms are at the edges of the unit cell,and $Na$ atoms are at the body center. The molecular formula of the compound is:

The number of unit cells present in $39 \ g$ of potassium if it crystallizes as a body-centered cubic $(BCC)$ structure is ($N = \text{Avogadro number}$,$\text{At. wt. of potassium} = 39$).