Question

Tungsten has a density of \[19.35{\text{ g c}}{{\text{m}}^{ - 3}}\]and length of the side of the unit cell is 316 pm. The unit cell has a body-centred unit cell. How many atoms of the element does 50 g of the element contain?

Answer 1

Hint: For this question we need to use the formula for density of lattice or unit cell. In a body centered unit cell the numbers of atoms are 2.
Formula used:
 \[{\text{density }} = \dfrac{{{\text{Atomic mass }} \times {\text{ Z}}}}{{{\text{Volume }} \times {\text{ }}{{\text{N}}_{\text{A}}}}}{\text{ }}\]
Here Z is the number of atoms in a unit cell and \[{{\text{N}}_{\text{A}}}\] is the Avogadro number or constant.

Complete step by step answer:
A body centered unit cell is the unit cell that contains 8 atoms on the corner and 1 atom in the body center. One atom at the corner is shared by 8 other atoms. So the contribution of one atom at a corner is one eighth. The atom present in the body center has the full contribution that is 1.
So the total number of atom will be calculated by multiplying the number of atoms present in one unit cell with their contribution as:
\[\dfrac{1}{8} \times 8 + 1 \times 1 = 2\]
That is \[{\text{Z}} = 2\].
The side of the unit cell is 316 pm. The volume of unit cell will be calculated as \[{(316{\text{ pm)}}^3}\]
\[{\text{1 pm }} = {10^{ - 10}}{\text{cm}}\]
Hence we will get \[{(316{\text{ pm)}}^3} = {(316 \times {10^{ - 10}}{\text{ cm)}}^3}\]
We will first calculate the atomic mass for the \[{{\text{N}}_{\text{A}}}\] atoms that is \[{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}\] atoms.
Now we will substitute the known variables in the formula:
\[19.35{\text{ g c}}{{\text{m}}^{ - 3}}{\text{ }} = \dfrac{{{\text{Atomic mass }} \times {\text{ 2}}}}{{{{(316{\text{ pm)}}}^3}{\text{ }} \times {\text{ 6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}}}{\text{ }}\]
Rearranging we will get
\[ \Rightarrow {\text{Atomic mass }} = \dfrac{{19.35{\text{ g c}}{{\text{m}}^{ - 3}} \times {{(316 \times {{10}^{ - 10}}{\text{ cm)}}}^3}{\text{ }} \times {\text{ 6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}}}{2}{\text{ }}\]
\[{\text{Atomic mass }} = 183.845{\text{ g}}\]
Now \[183.845{\text{ g}}\] of tungsten contains \[{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}\] atoms.
Hence 50 g of tungsten will contain,
\[ \Rightarrow \dfrac{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}}}{{183.845{\text{ g}}}} \times 50{\text{g}} = 1.637 \times {\text{1}}{{\text{0}}^{23}}{\text{ atoms}}\]

Hence the numbers of atoms are \[1.637 \times {\text{1}}{{\text{0}}^{23}}\]

Note:
A unit cell is the building block of a crustal. A crystal is formed by repeating lattice and a lattice is formed by repeating unit cells in different directions. There are various types of unit cell such as simple cubic unit cell, face centered unit cell, body centered unit cell, edge centered unit cell etc.