Question

An FCC lattice has lattice parameters ${\text{a}} = 400{\text{pm}}$. Calculate the molar volume of the lattice including all the empty space.
A.$10.8{\text{mL}}$
B.$96{\text{mL}}$
C.$8.6{\text{mL}}$
D.$9.6{\text{mL}}$

Answer 1

To solve this question, you must recall the formula used to find the number of atoms in a unit cell. In a Face Centred Cubic lattice (FCC), the atoms are present on the corners as well as at the centres of all faces.
Formula used:
$n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4}$
Where, n = effective number of atoms in unit cell
${n_c}$ is number of atoms at the corners of the unit cell
${n_f}$ is number of atoms at the six faces of the unit cell
${n_i}$is number of atoms completely inside the unit cell
${n_e}$ is number of atoms at the edge centres of the unit cell

Complete step by step answer:
In the question, we are supposed to find the volume of one mole of the face centred cubic lattice.
Firstly, we must find the volume of the unit cell. We have with us a cubic unit cell.
It should be known that Volume of a cube:
${\text{V}} = {{\text{a}}^3} = \left( {400 \times {{10}^{ - 10}}} \right)$
Since, we know that the effective number of atoms in a face centred lattice is four
Thus we can say that, the volume of four atoms is $64 \times {10^{ - 24}}{\text{c}}{{\text{m}}^3}$
One mole of a substance contains $6.023 \times {10^{23}}$ particles (atoms, or molecules, or ions)
So, the volume of $6.023 \times {10^{23}}$atoms is given by,
${\text{V'}} = \left( {64 \times {{10}^{ - 24}}{\text{c}}{{\text{m}}^3}} \right)\dfrac{{\left( {6.023 \times {{10}^{23}}} \right)}}{4}$
Therefore we get, ${\text{V}}' = 9.6{\text{c}}{{\text{m}}^3} = 9.6{\text{mL}}$

Thus, the correct option is D.

Note:
It should be known that in a face centred cubic unit cell, atoms are present at the corners and the face centres of the cube. So, we can write that,
${n_c} =$ number of atoms at the corners of the unit cell $= 8$
${n_f} =$number of atoms at the six faces of the unit cell $= 6$
${n_i} =$ number of atoms completely inside the unit cell $= 0$
${n_e} =$ number of atoms at the edge centres of the unit cell $= 0$
Thus, the total number of atoms in a face centred unit cell is
$n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4}$
Substituting the values, we get
$n = \dfrac{8}{8} + \dfrac{6}{2} + \dfrac{0}{1} + \dfrac{0}{4}{\text{ }}$
$n = 1 + 3 + 0 + 0{\text{ }}$
Thus, $n = 4$