Question

Silver crystallises in FCC structure. If the density of silver is $10.51g.c{{m}^{-3}}$. Calculate the volume of the unit cell. Atomic mass of silver (Ag) = $108 g{{m}^{-1}}$.

Answer 1

Hint: FCC is face centred cubic unit cell. It contains four atoms per unit cell and has a coordination number of 12. To calculate the volume of unit cell we will use the formula:
\[Volume\text{ }of\text{ }unit\text{ }cell=\frac{mass\text{ }of\text{ }unit\text{ }cell}{density\text{ }of\text{ }silver}\]
Formula used: \[Volume\text{ }of\text{ }unit\text{ }cell=\frac{mass\text{ }of\text{ }unit\text{ }cell}{density\text{ }of\text{ }silver}\]

Complete step by step answer:
- We are being provided with a density of Ag= $10.51 g.c{{m}^{-3}}$, and atomic mass of silver (Ag) = $108 g{{m}^{-1}}$.
- We have to find the volume of the unit cell, for which we will first calculate the mass of one atom of silver. We can use the following formula for this:
\[\frac{molar\text{ }mass\text{ }of\text{ }silver}{{{N}_{A}}}\]
${{N}_{A}}=6.022\times {{10}^{23}}$; ${{N}_{A}}$is the Avogadro number, value of${{N}_{A}}=6.022\times {{10}^{23}}$
\[\begin{align}
  & =\frac{108}{6.022\times {{10}^{23}}} \\
 & =17.93\times {{10}^{-23}} \\
\end{align}\]
- Now, we will find the mass of unit cell of silver = Atoms in unit cell x mass of one atom
So, by putting values in this formula we get,
\[\begin{align}
  & =4\times 17.93\times {{10}^{-23}} \\
 & =71.72\times {{10}^{-23}} \\
\end{align}\]
- So, as we know the value of mass of unit cell and the density of silver, we can calculate the volume of unit cell. As , \[Density=\frac{mass\text{ }of\text{ }unit\text{ }cell}{volume\text{ }of\text{ }unit\text{ }cell}\]
We can use the formula: \[Volume\text{ }of\text{ }unit\text{ }cell=\frac{mass\text{ }of\text{ }unit\text{ }cell}{density\text{ }of\text{ }silver}\]
By putting the values in this formula we get:
\[\begin{align}
  & Volume\text{ }of\text{ }unit\text{ }cell={{\frac{71.72\times 10}{10.51}}^{-23}} \\
 & = 68.24\times {{10}^{-23}}c{{m}^{3}} \\
\end{align}\]
Hence, we can conclude that the volume of the unit cell is $68.24\times {{10}^{-23}}c{{m}^{3}}$.

Note: We must not forget to write the unit after solving any question. We must not get confused in terms of FCC and BCC. FCC is face centred cubic unit cell, it contains four atoms per unit cell. Whereas, BCC is a body centred unit cell, it contains 2 atoms per unit cell.