Question

Aluminium (atomic mass$= 27$) crystallises in a cubic system with edge length of $4A$. Its density is $2.7gc{m^{ - 3}}$ . The number of aluminium atoms present per unit cell is:
A. $5$
B. $6$
C.$4$
D.$2$
E.$3$

Answer 1

Density: It is defined as the mass per volume. Hence density of unit cell is mass of unit cell divided by volume of unit cell. The mass of a unit cell is equal to the product of the number of atoms in a unit cell and mass of each unit cell.

Complete step by step answer:
Unit of crystal: It is defined as the repeating unit in the crystal.and there types are: (I) primitive cubic unit cell, (II) Body-centered cubic unit cell (III) Face-centered cubic unit cell.
Primitive cubic unit cell: In this unit cell, atoms are present only at the corners of the cube. We know that there are eight corners in a cube. So one atom is shared by right corners. So the contribution of an atom in one corner is $\dfrac{1}{8}$. And the number of atoms present in a primitive cubic unit cell is one.
Body-centered cubic unit cell:These are those types of unit cell in which, atoms are present at the centre and corners of the cube. We know that there is only one centre in a cube. So one atom is completely present there. And one atom is contributing to the eight corners. So the total number of atoms present in a Body-centered cubic unit cell is $2$.
Face-centered cubic unit cell: In this unit cell, atoms are present at corners and at the face centres of the cube. There are a total six faces in a cube. therefore one atom contributes to two face centres i.e. one atom is half. hence one atom contributing to all eight corners of the cube. So there are four atoms present in the face-centered cubic unit cell.
Density: It is defined as the mass per volume of the unit cell.
Here in the question we are given with the density of the unit cell and edge length of the unit cell. If we are given with the edge length we can calculate the volume of the unit cell i.e. if $a$ is the edge length then ${a^3}$ will be the volume of the unit cell. In the question $a = 4\mathop A\limits^ \circ = 4 \times {10^{ - 8}}cm$. So its volume will be ${a^3} = {(4 \times {10^{ - 8}})^3} = 64 \times {10^{ - 24}}c{m^3}$. The density is given as $2.7gc{m^{ - 3}}$. And mass is given as $27g/mol$.

Now we know that the density of a unit cell is the mass of the unit cell divided by the volume of the unit cell.
$d = \dfrac{{Z \times m}}{{{N_A}{a^3}}}$where ${N_A}$ is Avogadro number which is equal to $6.022 \times {10^{23}}$, $Z$ is number of atoms per unit cell, $m$ is mass and $a$ is edge length .
After putting the values of density, mass, volume, Avogadro number we will get the number of atoms per unit cell.
$Z = \dfrac{{d \times {N_A} \times {a^3}}}{m}$
$\Rightarrow Z = \dfrac{{2.7 \times 64 \times {{10}^{ - 24}} \times 6.02 \times {{10}^{23}}}}{{27}} = 3.85 \\\ Z \simeq 4 \\\$

Therefore, The number of atoms per unit cell in this question will be four. So option C is correct. .

Note: As the number of atoms per unit cell is four then the compound may be face centered cubic cell because in this cubic cell the atoms are present at the face center of the cubic cell and at the corners of the unit cell.