Question

Determine the density of cesium chloride which crystallizes in a bcc type structure with the edge length \[{\rm{412}}{\rm{.1}}\,{\rm{pm}}\]. The atomic masses of \[{\rm{Cs}}\] and \[{\rm{Cl}}\] are \[{\rm{133}}\] and \[{\rm{35}}{\rm{.5}}\] respectively. \[\left( {4.0\;{\rm{g}}\;{\rm{c}}{{\rm{m}}^{ - 3}}} \right)\]

Answer 1

Hint: We know that the number of atoms per unit cell in a crystal which crystallizes in a bcc system is two. The density of the atom in any of the system varies directly with the number of atoms in the unit cell of the system.

Complete step by step answer:
We know that the chemical formula of cesium chloride is \[{\rm{CsCl}}\]. Therefore, we can calculate the molar mass of it by adding the atomic masses of \[{\rm{Cs}}\] and \[{\rm{Cl}}\]. Then the molar mass of \[{\rm{CsCl}}\] is \[133 + 35.5 = 168.5\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\].
We can take the value edge length of the unit cell from the question which is \[{\rm{412}}{\rm{.1}}\,{\rm{pm}}\].
We know that, \[{\rm{1}}\,{\rm{pm}} = {10^{ - 10}}\;{\rm{cm}}\]. Therefore, we can convert the given value of lattice parameter in cm. therefore, we can say that the lattice parameter is equal to \[{\rm{412}}{\rm{.1}} \times {10^{ - 10}}\;{\rm{cm}}\].
The value of $Z = 1$ as there is one Cs and one Cl atom per unit cell.
We know that the formula used for calculating that is given by:
\[\rho = \dfrac{{{\rm{Z}} \times {\rm{M}}}}{{{a^3} \times {{\rm{N}}_{\rm{A}}}}}\]
Here, \[{\rm{Z}}\] is the number of atoms per unit cell, \[{\rm{M}}\] is the molar mass and \[a\] lattice parameter.
When we substitute all the values in the above equation we get the value of density as follows.
\[\begin{array}{c}
\rho = \dfrac{{{\rm{1}} \times 168.5\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}{{{{\left( {412.1 \times {{10}^{ - 10}}\;{\rm{cm}}} \right)}^3} \times 6.02 \times {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}\\
 = 4.0\;{\rm{g}}\;{\rm{c}}{{\rm{m}}^3}
\end{array}\]
Thus, we can say that the density of cesium chloride in the bcc system is \[4.0\;{\rm{g}}\;{\rm{c}}{{\rm{m}}^3}\].

Note:
We know that by knowing the unit-cell dimensions we can calculate the theoretical density of a cubic crystal. The theoretical density obtained using the formula used in the question is with the assumption that each lattice point is occupied by the species. But if some lattice points remain vacant, then we can also calculate the percentage occupancy from the observed and the theoretical densities