Question

A metal crystallizes face-centered cubic lattice with the edge length of $\,450pm\,$. Molar mass of the metal is $\,50gmo{l^{ - 1}}\,$. The density of metal will be:
A.$\,2.64g\,c{m^{ - 3}}\,$
B.$\,3.64g\,c{m^{ - 3}}\,$
C.$\,4.64g\,c{m^{ - 3}}\,$
D.$\,2.68g\,c{m^{ - 3}}\,$

Answer 1

Hint: The density of the unit cell can be found out using the respective formula with the given data Density is the ratio of mass and volume of the unit cell. The number of atoms in the unit cell of a face-centered cubic unit cell is equal to four. Hence, the equation varies for each unit cell based on this number.

Complete step by step answer:
Let us first understand the properties of a face centred unit cell;
Atoms at all corners of the crystal lattice and at the middle of all the faces of the cube are found in the FCC unit cell. The face-centered atom is divided between two neighbouring unit cells, and only half of each atom belongs to a single cell.
Now, coming to the question we have to find the density here;
In general,
$Density = \dfrac{{Mass}}{{Volume}}\,$
Same applied to the unit cell also.
However, the mass and volume here can be derived with certain values;
$\,Mass{\text{ }}of{\text{ }}unit{\text{ }}cell{\text{ }} = {\text{ }}number{\text{ }}of{\text{ }}atoms{\text{ }}in{\text{ }}unit{\text{ }}cell{\text{ }} \times {\text{ }}mass{\text{ }}of{\text{ }}each{\text{ }}atom{\text{ }} = {\text{ }}z{\text{ }} \times {\text{ }}m$
Where,
$\,z = \,$ number of atoms in the unit cell
$\,m = \,$ Mass of each atom
With the aid of the Avogadro number and molar mass, the mass of an atom can be given as;
$\,\dfrac{M}{{{N_A}}}\,$
Where, $\,M{\text{ }}$ is the molar mass
$\,{N_A}$ is the Avogadro’s number
Volume can be given as;
$\,V = {a^3}\,$
Where, $\,a\,$ is the edge length.
$\,\therefore Density\,of\,unit\,cell = \dfrac{m}{v} = \dfrac{{z \times m}}{{{a^3}}} = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}\,$
Here, in this face centred unit cell;
$\,z = 4\,$
$\,M = 50gmo{l^{ - 1}}\,$
$\,a = 450pm = 450 \times {10^{ - 10}}cm\,$
${N_A} = 6.022 \times {10^{23}}$$\,mol\,$
$\,\therefore Density = \dfrac{{4 \times 50}}{{{{(450 \times {{10}^{ - 10}})}^3} \times 6.022 \times {{10}^{23}}}}\,$
$\, = 3.64g\,c{m^{ - 3}}\,$

Hence, for this question, option B is the correct answer.

Note:
As there are four atoms present in the face centred unit cell, the density of elements would be four times larger as well. The face-centered cubic unit cell is the simplest repeating unit in a cubic closest-packed structure. In fact, in this structure, the existence of face-centered cubic unit cells itself explains why the structure is known as the closest-packed cubic structure because of the presence of octahedral holes within them.