Question

The unit cell length of $NaCl$ crystal is $564pm$ . Its density would be
$A.$ $1.082g/c{m^3}$
$B.$ $2.165g/c{m^3}$
$C.$ $3.247g/c{m^3}$
$D.$ $4.33g/c{m^3}$

Answer 1

As we know the structure of $NaCl$ crystal is face-centred cubic type cubic lattice. In this face-centred cubic arrangement, there is one extra atom at the centre of the six faces of the unit cube.

Complete step by step answer:
In the given question it is given that the unit cell length of $NaCl$ crystal is $564pm$. We have to determine the density. Before calculating its density we have to know the relation between unit cell length and density. Suppose the edge unit cell of a cubic crystal cubic determined by $X -$ ray diffraction is $a,d$ the density of the solid substance and $M$ the molar mass. In case of cubic crystal:
Volume of a unit cell$= {a^3}$
And, the mass of the unit cell is equal to the number of atoms in the unit cell $\times$ mass of each atom.
Now, we know the formula of density$= \dfrac{{{{Mass of unit cell}}}}{{{{Volume of unit cell}}}}$
$density\left( d \right) = \dfrac{{Z \times M}}{{{a^3} \times {N_A}}}$
Where $M$represent the mass of unit is cell and ${a^3}$ is the volume of unit cell, ${N_A}$ is the Avogadro’s number and $Z$ is the number of atoms present in one unit cell, in this case it is $4$.
Now, we know the relation between unit cell length and density.
$d = \dfrac{{Z \times M}}{{{a^3} \times {N_A}}}$
By putting the value of known quantity, we will get the density of $NaCl$ crystal.
$d = \dfrac{{4 \times 58.5g/mol}}{{\left( {5.64 \times {{10}^{ - 10}}cm} \right) \times 6.022 \times {{10}^{23}}}}$ $\left[ {M = 58.5g/mol} \right]$
On solving the above equation,
$d = 2.165g/c{m^3}$

So, the correct option is $B.$

Note:
Always remember that the density of a unit cell is the same as the density of a substance. We can determine the density of solid by other methods. Out of the five quantities $(a,M,Z,d\& {N_A})$ , if any four quantities are known then we can determine the fifth quantity.