Question

The number of atoms in $2.4{\text{ g}}$ of body centred cubic crystal with edge length $200{\text{ pm}}$is:
( Density$= 10{\text{ g c}}{{\text{m}}^{ - 3}}$, ${{\text{N}}_{\text{A}}} = 6 \times {10^{23{\text{ }}}}{\text{atoms/mol}}$)
A.$6 \times {10^{19}}$
B.$6 \times {10^{20}}$
C.$6 \times {10^{23}}$
D.$6 \times {10^{22}}$

Answer 1

First we must find the molecular mass of the given crystal using the available data. To find that, you must recall the formula for the density of a unit cell. We shall substitute the appropriate values in the formula given.
Formula used: $d = \dfrac{{n \times M}}{{{N_A} \times {a^3}}}$
Where, $d$ represents the density of the unit cell
$n$ denotes the number of atoms present in a unit cell
$M$ denotes the molecular mass of the given substance.
And, $a$ denotes the length of edge of the cubic unit cell

Complete step by step answer:
We know that, in a body centred cubic unit cell the atoms/ molecules are present at the corners of the cell and one at the centre of the cell. So, the number of atoms/ molecules present in a body centred cubic lattice is $n = 2$
Using the formula for density,
$d = \dfrac{{n \times M}}{{{N_A} \times {a^3}}}$
Rearranging:
$M = \dfrac{{d \times {N_A} \times {a^3}}}{n}$
Substituting the values, we get,
$M = \dfrac{{10 \times 6.022 \times {{10}^{23}} \times {{\left( {2 \times {{10}^{ - 8}}} \right)}^3}}}{2}$
$M = 24{\text{ g/mole}}$
Now we have the molecular mass of the given substance and so, we can find the number of atoms present easily by finding the number of moles in $2.4{\text{ g}}$ of the substance.
${\text{moles}} = \dfrac{{2.4}}{{24}} = 0.1{\text{ mole}}$
Thus, the number of atoms is given by $N = 0.1 \times {N_A} = 0.1 \times 6.022 \times {10^{23}}$
$N = 6.022 \times {10^{22}}$

Thus, the correct answer is D.

Note:
It should be known that in a body centred cubic unit cell, atoms are present at the corners and the centre of the cube. So, we can write that,
${n_c} =$ number of atoms present at the corners of the unit cell $= 8$
${n_f} =$number of atoms present at the six faces of the unit cell $= 0$
${n_i} =$ number of atoms present completely inside the unit cell $= 1$
${n_e} =$number of atoms present at the edge centres of the unit cell $= 0$
Thus, the total number of atoms in a body centred unit cell is
$n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4}$
Substituting the values, we get
$n = \dfrac{8}{8} + \dfrac{0}{2} + \dfrac{1}{1} + \dfrac{0}{4}{\text{ }}$
$n = 1 + 0 + 1 + 0{\text{ }}$
Thus, $n = 2$