An element having atomic mass 63.1 g/mol has a face centered cubic unit cell with edge length \[\; = 3.608 \times {10^{ - 8}}cm\], Calculate the density of the unit cell.
[Given\[{{\text{N}}_{\text{A}}} = 6.022 \times {10^{23}}\] ]

Hint:As for face centered cubic cell,
We have that Z = 4 where we know that, Z = number of atoms
Also, we know the given value of M \[ = 63.1{\text{ }}g/mol\] where, M = atomic mass
\[a = 3.608 \times {10^{ - 8}}cm\], where, a = edge length given value in question

Formula Used: Density, \[d = \dfrac{{Z \times N}}{{{a^3} \times {N_A}}}\]

Complete step by step solution:
Given value in the question are as below:-
Where Z = 4
Atomic mass is given as M = 63.1 g
While the edge length is given as \[a = 3.608 \times {10^{ - 8}}cm\]
As we know that,\[{{\text{N}}_{\text{A}}} = 6.022 \times {10^{23}}\]
For calculation of density we have,
\[d = \dfrac{{Z \times N}}{{{a^3} \times {N_A}}}\]
Putting the values of all the terms we will get,
$d = \dfrac{{4 \times 63.1}}{
  {(3.608 \times {10^{ - 8}})^3} \times 6.022 \times {10^{23}}} \\$
$ = 8.923g/c{m^3}
$

Hence, the value of density will be \[ = d = 8.923{\text{g/c}}{{\text{m}}^{\text{3}}}\]

Additional information:
Its international system unit is reciprocal mole, and known as \[{N_A}\; = {\text{ }}6.02214076 \times {10^{23}}\;mo{l^{ - 1}}.\] It’s name is kept as of Italian scientist Amedeo Avogardro so it is called as Avogadro constant he was not the chemist who discovered it.For FCC cell the sum of atomic number (Z) is always 4.
Avogadro’s number ($N_A$) is always \[6.022 \times {10^{23}}Atoms/Mole\]
1. Primitive unit cell the sum of atoms in a unit cell, z is equal to 1. So the, density is given as:
Density of unit cell = \[\dfrac{{1 \times M}}{{{a^3} \times {N_A}}}\]
2. BCC unit cell the sum of atoms in a unit cell, z is equal to 2. Hence, density is given as:
Density of unit cell = \[\dfrac{{2 \times M}}{{{a^3} \times {N_A}}}\]
3. FCC unit cell the sum of atoms in a unit cell, z is equal to four. Hence, density of unit cell is given as:
Density of unit cell = \[\dfrac{{3 \times M}}{{{a^3} \times {N_A}}}\]

Note:
Take note of the SI unit of density i.e. g/cm3,
And the value of the Avogadro’s number which is always predefined as,
\[N_A=6.022 \times {10^{23}}Atoms/Mole\]
Hence, \[{\text{Density of Unit Cell = }}\dfrac{{{\text{Mass of Unit Cell}}}}{{{\text{Volume of Unit Cell}}}}\]