At 100$$^{\circ}$$C, copper (Cu) has FCC unit cell structure with cell edge length of $$x$$ $$\text{Å}$$. What is the approximate density of Cu (in g cm$$^{-3}$$) at this temperature? [Atomic Mass of Cu = 63.55 u]

For any crystalline solid, the density $$\rho$$ is obtained from the general relation

$$\rho=\dfrac{Z\,M}{N_A\,a^{3}}$$

where

$$Z=$$ number of atoms in one unit cell,

$$M=$$ molar (atomic) mass in g mol−1,

$$N_A=$$ Avogadro constant, and

$$a=$$ edge length of the cubic unit cell in cm.

We are told that copper crystallises in the face-centred cubic (FCC) structure at 100 $$^{\circ}$$C. In an FCC unit cell we have

$$Z=4$$

The atomic mass of copper is given as

$$M = 63.55\ \text{g mol}^{-1}$$

The edge length is quoted as $$x\ \text{Å}$$. Because our density must come out in g cm−3, we first convert the edge length to centimetres. The conversion factor is

$$1\ \text{Å}=1\times10^{-8}\ \text{cm}$$

So

$$a = x\ \text{Å}=x\times10^{-8}\,\text{cm}$$

Now we substitute every symbol into the density formula:

$$\rho=\dfrac{Z\,M}{N_A\,a^{3}}=\dfrac{4\;\times\;63.55}{\;6.022\times10^{23}\; \times\;(x\times10^{-8})^{3}}\ \text{g cm}^{-3}$$

Let us treat the edge length factor first:

$$(x\times10^{-8})^{3}=x^{3}\times(10^{-8})^{3}=x^{3}\times10^{-24}$$

Inserting this result gives

$$\rho=\dfrac{4\times63.55}{6.022\times10^{23}\times x^{3}\times10^{-24}}\ \text{g cm}^{-3}$$

We now focus on the powers of ten. The product $$6.022\times10^{23}\times10^{-24}=6.022\times10^{-1}=0.6022$$. Hence

$$\rho=\dfrac{4\times63.55}{0.6022\,x^{3}}\ \text{g cm}^{-3}$$

The numerator multiplies to

$$4\times63.55=254.20$$

Therefore

$$\rho=\dfrac{254.20}{0.6022\,x^{3}}\ \text{g cm}^{-3}$$

Carrying out the division,

$$\dfrac{254.20}{0.6022}\approx422$$

So we finally have

$$\rho\approx\dfrac{422}{x^{3}}\ \text{g cm}^{-3}$$

This matches the expression offered in Option C.

Hence, the correct answer is Option C.