Answer

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**Hint:**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, $n = 2$

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