A metal (atomic mass = 50) has a body centred cubic crystal structure. The density of the metal is 5.96 g `cm^(-3)`. Find the volume of this unit cell.

Explanation

To find the volume of the unit cell of a metal with a body-centered cubic (BCC) structure, we can use the relationship between density, mass, and volume. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Atomic mass (m) = 50 g/mol
- Density (D) = 5.96 g/cm³
- Avogadro's number (Nₐ) = 6.02 x 10²³ atoms/mol
- For BCC structure, the number of atoms per unit cell (z) = 2

### Step 2: Use the formula for density
The formula for density (D) is given by:
\[ D = \frac{z \cdot m}{V} \]
where:
- \(D\) = density
- \(z\) = number of atoms per unit cell
- \(m\) = molar mass (mass of one mole of atoms)
- \(V\) = volume of the unit cell

### Step 3: Rearrange the formula to find the volume
We can rearrange the formula to solve for the volume \(V\):
\[ V = \frac{z \cdot m}{D} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ V = \frac{2 \cdot 50 \text{ g/mol}}{5.96 \text{ g/cm}^3} \]

### Step 5: Calculate the volume
Calculating the volume:
\[ V = \frac{100 \text{ g/mol}}{5.96 \text{ g/cm}^3} \approx 16.78 \text{ cm}^3/\text{mol} \]

### Step 6: Convert the volume to the volume of the unit cell
To find the volume of the unit cell in cubic centimeters, we need to divide the volume per mole by Avogadro's number:
\[ V_{\text{unit cell}} = \frac{16.78 \text{ cm}^3/\text{mol}}{6.02 \times 10^{23} \text{ atoms/mol}} \approx 2.78 \times 10^{-23} \text{ cm}^3 \]

### Final Answer
The volume of the unit cell is approximately \(2.78 \times 10^{-23} \text{ cm}^3\).

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