A metal (atomic mass = 40) has a body centred cubic crystal structure. If the density of the metal is 4.50 g `cm^(-3)`, calculate the volume of the unit cell.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Atomic mass (m) = 40 g/mol
- Density (ρ) = 4.50 g/cm³
- For a body-centered cubic (BCC) structure, the number of atoms per unit cell (Z) = 2
- Avogadro's number (NA) = 6.022 x 10²³ atoms/mol

### Step 2: Write the formula for density
The formula for density in terms of the unit cell volume is given by:
\[ \rho = \frac{Z \cdot m}{V} \]
where V is the volume of the unit cell.

### Step 3: Rearrange the formula to find the volume of the unit cell
We can rearrange the density formula to solve for the volume (V):
\[ V = \frac{Z \cdot m}{\rho} \]

### Step 4: Substitute the values into the equation
Now we substitute the known values into the equation:
\[ V = \frac{2 \cdot 40 \, \text{g/mol}}{4.50 \, \text{g/cm}^3} \]

### Step 5: Calculate the volume
Calculating the volume:
\[ V = \frac{80 \, \text{g/mol}}{4.50 \, \text{g/cm}^3} = 17.78 \, \text{cm}^3/\text{mol} \]

### Step 6: Convert the volume from cm³/mol to cm³
To find the volume of the unit cell in cm³, we need to convert the volume from per mole to per unit cell using Avogadro's number:
\[ V_{\text{unit cell}} = \frac{17.78 \, \text{cm}^3/\text{mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \]

Calculating this gives:
\[ V_{\text{unit cell}} = 2.95 \times 10^{-23} \, \text{cm}^3 \]

### Final Answer
The volume of the unit cell is:
\[ V_{\text{unit cell}} = 2.95 \times 10^{-23} \, \text{cm}^3 \]

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