Question

Aluminum crystallizes in a cubic close-packed structure. Its metallic radius is 125 pm. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in $1.00{\mathrm{cm}}^3$ of aluminum?

Answer 1

Step 1: Given data

The radius of the metal is $125\mathrm{pm}$

Volume of Aluminium given is $1.00{\mathrm{cm}}^3$

Step 2: Calculating the length

We know that for closed-packed structures:

$$\begin{gathered} \mathrm{a}=2\sqrt{2}\mathrm{r} \\ \Rightarrow \mathrm{a}=2\sqrt{2}\times 125 \\ \Rightarrow \mathrm{a}=354\mathrm{pm} \end{gathered}$$

Step 3: Calculating the number of unit cell

As we have the radius (r) as $125\mathrm{pm}$ and the length (a) as $354\mathrm{pm}$

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{unit}\mathrm{cell}={\mathrm{a}}^3 \\ \mathrm{V}={\left(354\mathrm{pm}\right)}^3 \\ \mathrm{V}=4.4\times {10}^{-23}{\mathrm{cm}}^3[\because 1\mathrm{pm}={10}^{-10}\mathrm{cm}] \end{gathered}$$

$\therefore$Number of unit cells in $1.00{\mathrm{cm}}^3$ of aluminium is:

$\frac{1.00}{4.4\times {10}^{-23}}=2.27\times {10}^{22}$

Therefore, length of the side of the unit cell is 354 pm and Number of unit cells in $1.00{\mathrm{cm}}^3$ of aluminium is $2.27\times {10}^{22}$