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An element ‘X’ has an f.c.c unit cell. Therefore, the number of atoms present per unit, Z = 4.

Given, atomic mass of the element, M = 40g $mo{{l}^{-1}}$

If ‘a’ is the edge length of a unit cell. Then, the volume of the unit cell will be equal to ‘${{\text{a}}^{3}}$’

Given, edge length of unit cell of element ‘X’ having f.c.c structure, a = 400 pm = 400 $\times {{10}^{-12}}$ m.

Density of a unit cell can be given as, d = $\dfrac{\text{mass of unit cell}}{\text{volume of unit cell}}$

Mass of the unit cell will be equal to the total number of atoms present in the unit cell multiplied by the mass of each atom, i.e. m$\times $ Z. For f.c.c unit cell, Z = 4.

Therefore, the density of the unit cell can be calculated as\[\begin{align}

& d=\dfrac{m\times Z}{V} \\

& =\dfrac{m\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}} \\

& \because m\times {{N}_{A}}=M\Rightarrow m=\dfrac{M}{{{N}_{A}}} \\

& \Rightarrow d=\dfrac{M\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times {{N}_{A}}} \\

& d=\dfrac{40g\times 4}{{{(400\times {{10}^{-12}}m)}^{3}}\times 6.022\times {{10}^{23}}mo{{l}^{-1}}}=4.2\times {{10}^{-6}}g\,{{m}^{-3}} \\

\end{align}\]

Therefore, the density of the unit cell is 4.2$\times {{10}^{-6}}$ g${{m}^{-3}}$.

One mole of the element ‘X’ has 6.022$\times {{10}^{23}}$ atoms.

40 g of ‘X’ = 6.022$\times {{10}^{23}}$ atoms.

1 g of ‘X’ = $\dfrac{6.022\times {{10}^{23}}}{40}$ atoms

4 g of ‘X’ contains = 4$\times \dfrac{6.022\times {{10}^{23}}}{40}=6.022\times {{10}^{22}}$ atoms

We know that in an f.c.c unit cell one unit cell has 4 atoms.

1 unit cell = 4 atoms

1 atom = $\dfrac{1}{4}$ unit cell

Then, 6.022$\times {{10}^{22}}$ atoms = $\dfrac{1}{4}\times 6.022\times {{10}^{22}}$ = $1.505\times {{10}^{22}}$ unit cell.