An element ‘X’ (At. Mass = 40 g/mol ) having f.c.c structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4g of ‘X’. (N_A = 6.022 X 10²³ mol).

Here, $\mathrm{z}=4,\mathrm{a}=400\mathrm{pm}=4\times {10}^{-8}\mathrm{cm},\mathrm{M}=40$
As we know that $\mathbf{d}=\frac{\mathrm{Z}\mathbf{M}}{{\mathbf{a}}^3{\mathrm{N}}_{\mathrm{A}}}$
$\begin{array}{l}\mathrm{d}=\frac{4\times 40}{{\left(4\times {10}^{-8}\right)}^3\times 6.022\times {10}^{23}}\\ =\frac{160}{64\times {10}^{-24}\times 6.022\times {10}^{23}}=\frac{160}{64\times {10}^{-1}\times 6.022}\\ =\frac{160}{38.5408}=4.1514\approx 4.15\mathrm{g}{\mathrm{cm}}^{-3}\end{array}$
$40\mathrm{g}$ contains N_A atoms
4 g will  contain $\frac{{\mathrm{N}}_{\mathrm{A}}\times 4}{40}$ atoms
(FCC = 4 unit cells)
So, number of unit cell
$\begin{array}{l}=\frac{6.022\times {10}^{23}\times 4}{40}\times \frac{1}{4}\\ =\frac{6.022\times {10}^{22}}{4}=1.50\times {10}^{22}\text{ unit cell }\end{array}$