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

If a metal crystallises in a face-centred cubic $(FCC)$ structure with a metallic radius of $25 \ pm$, the number of unit cells in $1.0 \ cm^3$ of the lattice is:

Metal $M$ crystallizes in $fcc$ lattice. If the edge length of the unit cell is $4.077 \times 10^{-8} \ cm$ and the density is $10.5 \ g \ cm^{-3}$,the atomic mass of the metal is:

How many atoms of niobium are present in $2.43 \ g$ if it forms $bcc$ structure with density $9 \ g \ cm^{-3}$ and volume of unit cell $2.7 \times 10^{-23} \ cm^3$?

The density of $Li$ metal is $0.53 \ g/cm^3$ and the edge length of its unit cell is $3.5 \ \mathop A\limits^o$. Calculate the number of $Li$ atoms in the unit cell. $(N_A = 6.02 \times 10^{23} \ mol^{-1}, M = 6.94 \ g \ mol^{-1})$

Calculate the number of atoms present in a unit cell of an element having molar mass $190 \ g \ mol^{-1}$ and density $20 \ g \ cm^{-3}$. Given that $[a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}]$.