Calculate the number of unit cells in $10.8 \ g$ of metal,given that $\rho a^3 = 7.2 \times 10^{-22} \ g$.

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: $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$

Niobium crystallises in a body-centred cubic $(bcc)$ structure. If the density is $8.55 \, g \, cm^{-3}$,calculate the atomic radius of niobium using its atomic mass $93 \, u$.

$A$ compound forms a $bcc$ unit cell with an edge length of $400 \text{ pm}$. Determine the molar mass of the compound if the density of the compound is $3.5 \text{ g cm}^{-3}$. (Given: $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$)

$A$ metal $(X)$ of atomic weight $M \ g \ mol^{-1}$ crystallizes in a $bcc$ lattice. Its density is $d \ g \ cm^{-3}$. What is the equation for the unit cell edge length $(a)$? ($N=$ Avogadro number)

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: