Question

When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 408.7 pm. Calculate the density of silver in g/cm3?

Answer 1

The density of silver can be calculated by using the following formula:

$d=\frac{m}{V}$

Here, m is mass and V is volume.

To calculate density, mass of silver needs to be calculated first.

In FCC, the number of Ag atoms in a unit cell will be 4, atomic mass of Ag is 107.87 g/mol and number of atoms in 1 mol are $6.023\times 10^{23}$ atoms, thus, mass can be calculated as follows:

$m=4 atoms\times \frac{1 mol}{6.023\times 10^{23}atoms}\times \frac{107.87 g}{mol}=7.163\times 10^{-22} g$.

Now, volume can be calculated as follows:

$V=a^{3}$

First convert pm to cm:

$1 pm=10^{-10} cm$

Thus,

$408.7 pm=4.087\times 10^{-8} cm$

Putting the value to calculate volume,

$V=(4.087\times 10^{-8} cm)^{3}=6.83\times 10^{-23} cm^{3}$

Now, calculate density as follows:

$d=\frac{7.163\times 10^{-22}g}{6.627\times 10^{-23}cm^{3}}=10.5 g/cm^{3}$

Therefore, density of silver is $10.5 g/cm^{3}$.