Question

An element crystallizes in bcc structures. The edge length of its unit cell is 288pm. If the density of the crystal is $7.2gc{m^{ - 3}}$. What is the atomic mass ( in $g/mol$) of the element?
A. $51.8$
B. $103.6$
C. $25.9$
D. $207.2$

Answer 1

Hint: we know that bcc stands for body centred cube. It has eight atoms located at eight corners of the cube and an atom located at the centre of the cube. Every atom at the corners contributes $\dfrac{1}{8}$to the molecule and every atom at the centre contributes $1$ to the molecule. So the number of atoms in a body centred cube will be $(\dfrac{1}{8} \times 8) + 1 = 2$.

Complete step by step solution:
We already know that the number of atoms in a body centred cube is $2$. It is important to us to remember that the relation between the edge length and atomic radius will be $\sqrt 3 a = 4R$. Here $a = $edge length and $R = $ radius of the atom. The density of a cubic cell is given by $ \rho = \dfrac{{ZM}}{{{N_A}{A^3}}}$.
Here, $\rho = $density , $Z = $ number of atoms , ${N_A} = $Avogadro number , $A = $edge length and $M = $atomic mass of the element. We are given the value of $A = 288pm$, $\rho = 7.2gc{m^{ - 3}}$, ${N_A} = 6.022 \times {10^{23}}$ and $Z = 2$. Putting the values in the formula $ \Rightarrow \rho = \dfrac{{ZM}}{{{N_A}{A^3}}}$ we get
$
   \Rightarrow 7.2gc{m^{ - 3}} = \dfrac{{2M}}{{6.022 \times {{10}^{23}} \times {{(288 \times {{10}^{ - 10}})}^3}}} \\
   \Rightarrow M = \dfrac{{6.022 \times {{10}^{23}} \times 7.2 \times {{(288 \times {{10}^{ - 10}})}^3}}}{2} \\
   \Rightarrow M = 51.8g/mol \\
 $
So from the above explanation and calculation it is clear to us that

The correct answer of the given question is option: A.

Additional information:
We know that when a cube has eight atoms in eight corners and one atom in the body centre it is known as body centred cube. When a cube has eight atoms in eight corners and six atoms in the six faces of the cube it is known as a face centred cube.

Note: Always remember that the density of a crystal structure is given by the formula
$ \Rightarrow \rho = \dfrac{{ZM}}{{{N_A}{A^3}}}$.
Here, $\rho = $density , $Z = $ number of atoms , ${N_A} = $Avogadro number , $A = $edge length and $M = $atomic mass of the element. Always try to avoid calculation mistakes.