Question

Iron crystallizes in a bcc system with a lattice parameter of $2.861A^\circ$ . Calculate the density of iron in the bcc system (atomic weight of $Fe = 56$ , Avogadro’s no. $= 6.02 \times {10^{23}}mo{l^{ - 1}}$ )
A) $7.92gm{L^{ - 1}}$
B) $8.96gm{L^{ - 1}}$
C) $2.78gm{L^{ - 1}}$
D) $6.72gm{L^{ - 1}}$

Answer 1

Density in general is the ratio of mass of atom to its total volume in space. Mass of each atom is the ratio of molar mass to Avogadro’s no and the volume of a cube if the side is $a$ can be given by ${a^3}$ . Body centered cubic lattice has a contribution of $2$ .
Formula used: $d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$

Complete step by step solution
In the body centered cubic lattice the atoms are present in the eight corners like in simple cubic lattice and one extra atom is situated at the centre of body. Hence as the contribution of corner atoms are $\dfrac{1}{8}$ and the contribution of body atom is $1$ , the contribution of bcc will be
\therefore Z = 8(corneratom) \times \dfrac{1}{8} + 1(bodyatom) \times 1 \\\ \Rightarrow Z = 2 \\\
The packing efficiency of a solid is the percentage of total space filled by the particles. In bcc packing efficiency is considered to be $68\%$ .
Coordination number is the maximum number of atoms surrounding a particular atom in a lattice, and so the coordination number of bcc is $8$ .
Density of a solid can be determined by the formula -
$d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$ Where the density of solid is $d$ , $Z$ is the contribution of body centered cubic lattice, $M$ is the molar mass of element and ${N_A}$ is the Avogadro’s number.
In the mentioned question, molar mass of $Fe$ is $M = 56gmmo{l^{ - 1}}$ , lattice parameter $a = 2.861A^\circ$ , Avogadro’s no. ${N_A} = 6.02 \times {10^{23}}mo{l^{ - 1}}$ And the lattice mentioned is bcc which has a contribution of $Z = 2$
We have to convert the unit of lattice parameter from ${A^\circ }$ to $cm$ .
$\therefore a = 2.861 \times {10^{ - 8}}cm$
On substituting the values in the formula of density we get
$d = \dfrac{{2 \times 56gmo{l^{ - 1}}}}{{6.02 \times {{10}^{23}}mo{l^{ - 1}} \times {{(2.861 \times {{10}^{ - 8}}cm)}^3}}}$
Calculating the values,
$d = \dfrac{{112g}}{{6.02 \times {{10}^{23}} \times 23.418 \times {{10}^{ - 24}}c{m^3}}}$
On simplifying the equation we get
d = \dfrac{{112g}}{{14.097c{m^3}}} \\\ \Rightarrow d = 7.92gc{m^{ - 3}} \\\
Also $1c{m^3} = 1mL$
Hence the density of iron which crystallizes in bcc was found to be $7.92gm{L^{ - 1}}$ .
Therefore, the correct option is (A).

Note
While calculating the density it is very important to check the units of the value needed without unit conversion the value calculated can be considered as a wrong value. Like here we had changed the unit of lattice parameter from Armstrong into centimetres.