Question

An element crystallises in a bcc lattice with cell edge of $ 500pm $. The density of the element is $ 7.5gc{m^{ - 3}} $. How many atoms are present in $ 300g $ of the element?

Answer 1

Hint: We know that by knowing the dimensions of a unit cell, we can calculate the volume of the unit cell, similarly by knowing the density of a unit cell we can calculate the mass and number of atoms in the unit cell.

Complete answer:
As we know that each unit cell is adjacent to another and most of the atoms are shared by neighbouring unit cells due to which only some portion of each atom belongs to a particular unit cell.
So, by knowing the dimensions of a unit cell, we can calculate the volume of the unit cell, similarly by knowing the density of a unit cell we can calculate the mass and number of atoms in the unit cell.
Let the edge length of a cubic crystal of an element or compound $ = a\;cm $
Therefore the volume of the unit cell $ = {a^3}\;c{m^3} $
And density of the cell will be $ d = \dfrac{{mass}}{{volume}} $
Mass of a unit cell is given as $ mass = number\;of\;atoms\;in\;unit\;cell \times mass\;of\;each\;atom $
\[mass = Z \times m\], where m can be calculated as the ratio of atomic mass to the Avogadro’s number.
 $
 m = \dfrac{{atomic\;mass}}{{Avogadro's\;number}} \\
\Rightarrow m = \dfrac{M}{{{N_A}}} \\
 $
And mass of a unit cell $ = Z \times \dfrac{M}{{{N_A}}} $
Now using this we can find out the density:
 $ d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}} $
In the given question we are provided density $ = 7.5gc{m^{ - 1}} $, edge length = $ 500pm $ $ = 5 \times {10^{ - 8}}cm $ and mass = $ 300g $ so calculating the volume of the unit cell:
 $
\Rightarrow {a^3} = {(5 \times {10^{ - 8}})^3} \\
\Rightarrow {a^3} = 1.25 \times {10^{ - 22}}c{m^3} \\
 $
And we know that for a body centred cubic unit cell the number of atoms per unit cell is $ 2 $.
Put all the calculated and given values in the above formula:
 $ d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}} $
 $
\Rightarrow M = \dfrac{{d \times {N_A} \times {a^3}}}{Z} \\
\Rightarrow M = \dfrac{{7.5 \times 6.022 \times {{10}^{23}} \times (1.25 \times {{10}^{ - 22}})}}{2} \\
\Rightarrow M = 282.3g \\
 $
Now we know that the number of atoms can be calculated using the ratio of mass and Avogadro’s number to the molecular mass. So we get:
 $
 number\;of\;atoms = \dfrac{{mass \times {N_A}}}{M} \\
\Rightarrow no.of\;atoms = \dfrac{{300 \times 6.022 \times {{10}^{23}}}}{{282.3}} \\
\Rightarrow no.of\;atoms = 6.4 \times {10^{23}} \\
 $

Therefore the number of atoms present in $ 300g $ of elements is $ 6.4 \times {10^{23}} $ atoms.

Note:
the body centred cubic arrangement contains $ 8 $ lattice point at the corners and each one of these is shared by $ 8 $ cells and another lattice point is completely inside the unit cell or present at the centre of the body thus the number of points or atoms per unit cell of bcc is $ 8 \times \dfrac{1}{8} + 1 = 2 $ therefore the value of $ Z = 2 $ similarly value of ccp arrangement can be calculated which comes out to be $ Z = 4 $.