Find the radius of an atom in $fcc$ unit cell having edge length $405 \ pm$. (in $pm$)

Polonium (atomic mass = $209$) crystallises in a simple cubic structure with a density of $9.32 \ g \ cm^{-3}$. Its lattice parameter (in $pm$) is closest to:

An element crystallises in an $fcc$ unit cell with a cell edge length of $3.608 \times 10^{-8} \text{ cm}$. The density of the element is $8.92 \text{ g cm}^{-3}$. Calculate the atomic mass of the element $(N_A = 6.022 \times 10^{23} \text{ mol}^{-1})$. (in $\text{ g/mol}$)

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the nature of the cubic unit cell is: $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$

An element $X$ (At. wt. $= 80 \ g/mol$) has an $fcc$ structure. Calculate the number of unit cells in $8 \ g$ of $X$.

Calculate the volume of the unit cell for an element having a molar mass of $92 \ g \ mol^{-1}$ that forms a $bcc$ structure,given $\left[\varrho \times N_{A} = 5.0 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$.