Chromium metal (atomic mass = 52) has body centred cubic structure. The radius of chromium atom is 124.3 pm. Calculate the density of chromium metal.

Explanation

To calculate the density of chromium metal with a body-centered cubic (BCC) structure, we will follow these steps:

### Step 1: Identify the parameters
- Atomic mass (M) of chromium = 52 g/mol
- Radius (R) of chromium atom = 124.3 pm = \(124.3 \times 10^{-12}\) m
- Body-centered cubic structure (BCC) has Z = 2 (2 atoms per unit cell).

### Step 2: Calculate the edge length (A) of the unit cell
For a BCC structure, the relationship between the radius (R) and the edge length (A) is given by:
\[ A = \frac{4R}{\sqrt{3}} \]
Substituting the value of R:
\[ A = \frac{4 \times 124.3 \times 10^{-12}}{\sqrt{3}} = \frac{497.2 \times 10^{-12}}{1.732} \approx 287.07 \times 10^{-12} \text{ m} \]

### Step 3: Calculate the volume of the unit cell (A³)
Now, we calculate the volume of the unit cell:
\[ A^3 = (287.07 \times 10^{-12})^3 \approx 2.36 \times 10^{-33} \text{ m}^3 \]

### Step 4: Convert the volume to cm³
To convert from m³ to cm³, we use the conversion factor \(1 \text{ m}^3 = 10^6 \text{ cm}^3\):
\[ A^3 \approx 2.36 \times 10^{-33} \text{ m}^3 \times 10^6 \approx 2.36 \times 10^{-27} \text{ cm}^3 \]

### Step 5: Use the density formula
The formula for density (\(D\)) is given by:
\[ D = \frac{Z \times M}{A^3 \times N_A} \]
Where:
- \(Z = 2\) (number of atoms per unit cell)
- \(M = 52 \text{ g/mol}\)
- \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\) (Avogadro's number)

Substituting the values:
\[ D = \frac{2 \times 52}{2.36 \times 10^{-27} \times 6.022 \times 10^{23}} \]

### Step 6: Calculate the density
Calculating the denominator:
\[ 2.36 \times 10^{-27} \times 6.022 \times 10^{23} \approx 1.42 \times 10^{-3} \text{ g} \]
Now substituting back into the density formula:
\[ D \approx \frac{104}{1.42 \times 10^{-3}} \approx 7.30 \text{ g/cm}^3 \]

### Final Answer
The density of chromium metal is approximately **7.30 g/cm³**.

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