Question

An element crystallizes in a structure having FCC unit cells of an edge length 200 pm. Calculate the density (in $g\,c{m^{ - 3}}$) if 200 g of this element contains $24 \times {10^{23}}\,atoms$
A.$10.4$
B.$2.6$
C.$41.6$
D.$83.2$

Answer 1

Hint: We know that in a face centered cubic cell, the number of atoms present in unit cells is four. We can calculate the density using the number of atoms in the unit cell, mass of the element, Avogadro’s number and volume of the unit cell.
Formula used: We can calculate the density using the formula,
$Density = \dfrac{{z \times M}}{{N \times {{\left( a \right)}^3}}}$
Here, D=Density
z=Number of atoms in a unit cell
N=Avogadro’s number of atoms
V=Volume

Complete step by step answer:We know that an FCC unit cell comprises atoms at all the corners of the crystal lattice and atoms present at the centre of all the faces of the cube. The atom seen at the face-centered is shared between 2 adjacent unit cells and only 1/2 of each atom belongs to an individual cell.
In a face centered cubic unit cell, atoms are found in all corners of the lattice.
An atom is also present at the center of every face.
Two adjacent unit-cells are shared by a face-center atom.
Twelve atoms per unit cell.
Given data contains,
Mass of the element is 200g.
Edge length is 200pm.
Number of atoms is $24 \times {10^{23}}$.
Since they have given edge length, we have to calculate the volume using edge length.
We know the formula of volume as,
$V = {a^3}$
Here a=edge length
$1\mathop A\limits^ \circ = {10^{ - 8}}\,cm$
Let us now calculate the volume by substituting the value of edge length,
$
  V = {\left( {2 \times {{10}^{ - 8}}\,cm} \right)^3} \\
  V = 8 \times {10^{ - 24}}\,c{m^3} \\
$
We have calculated the volume as $8 \times {10^{ - 24}}\,c{m^3}$.
We know the number of atoms present in FCC unit cells is four.
We can calculate the density as,
$
  Density = \dfrac{{z \times M}}{{N \times {{\left( a \right)}^3}}} \\
  Density = \dfrac{{4 \times 200\,g}}{{24 \times {{10}^{23}} \times 8 \times {{10}^{ - 24}}\,c{m^3}}} \\
  Density = 41.67\,g\,c{m^{ - 3}} \\
$
We have calculated the density as $41.67\,g\,c{m^{ - 3}}$.

And hence Option (C) is correct.


Note:
An example of a compound that has face-centered cubic lattice is sodium chloride. Some other compounds that have face-centered cubic structures are lithium fluoride, lithium chloride, Sodium fluoride, potassium fluoride, potassium chloride etc. Some of the other crystal structures are Simple cubic cell, and body centered cubic cell.