Question

Aluminum has an atomic radius of 143 pm and forms a solid with a cubic closest packed structure. Calculate the density of solid aluminum in g/cm³.

Short answer

The density of aluminum is $2.72g/c{m}^3.$

Step-by-step solution

Finding l value

Use the Pythagoras theorem to obtain the l value as

$l=\surd \mathbf{8}\mathbf{r}$

From the given question:

r= 143pm,

So l is:

$$\begin{gathered} l=\sqrt{8}x\left(143\right)pm \\ =4.04x{10}^{-8}cm \end{gathered}$$

Volume of unit cell

$$\begin{gathered} Volume of one unitcell=l^3 \\ =(4.04x{10}^{-8}cm{)}^3 \\ =65.94x{10}^{-24}c{m}^3 \end{gathered}$$

density of aluminum

The mass of unit cell aluminum is $17.92x{10}^{-23}g$

The volume of unit cell aluminum is$65.94x{10}^{-24}c{m}^3$

$$\begin{gathered} Density=\frac{Mass}{volume} \\ =\frac{17.92x{10}^{-23}g}{65.94x{10}^{-24}c{m}^3} \\ =2.72g/c{m}^3 \end{gathered}$$

Thus the density of the aluminum is 2.72g/cm³