Determine the type of cubic lattice to which the iron crystal belongs if its unit cell has an edge length of 286 pm and the density of iron crystals is 7.86g `cm^(-3)`.

Explanation

To determine the type of cubic lattice to which the iron crystal belongs, we will follow these steps:

### Step 1: Gather Given Information
- Edge length of the unit cell (a) = 286 pm = \(286 \times 10^{-10}\) cm
- Density of iron (D) = 7.86 g/cm³
- Molar mass of iron (M) = 56 g/mol
- Avogadro's number (Nₐ) = \(6.022 \times 10^{23}\) mol⁻¹

### Step 2: Convert Edge Length to Appropriate Units
Since the density is given in g/cm³, we need to convert the edge length from picometers to centimeters:
\[ a = 286 \text{ pm} = 286 \times 10^{-10} \text{ cm} \]

### Step 3: Use the Density Formula
The formula for density (D) in terms of the number of formula units per unit cell (Z), molar mass (M), edge length (a), and Avogadro's number (Nₐ) is given by:
\[ D = \frac{Z \cdot M}{a^3 \cdot N_a} \]

### Step 4: Rearrange the Formula to Solve for Z
Rearranging the formula to find Z:
\[ Z = \frac{D \cdot a^3 \cdot N_a}{M} \]

### Step 5: Substitute the Values
Now, we will substitute the values into the equation:
\[ Z = \frac{7.86 \, \text{g/cm}^3 \cdot (286 \times 10^{-10} \, \text{cm})^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{56 \, \text{g/mol}} \]

### Step 6: Calculate \(a^3\)
First, calculate \(a^3\):
\[ a^3 = (286 \times 10^{-10})^3 = 2.33 \times 10^{-29} \, \text{cm}^3 \]

### Step 7: Calculate Z
Now substituting \(a^3\) back into the equation:
\[ Z = \frac{7.86 \cdot 2.33 \times 10^{-29} \cdot 6.022 \times 10^{23}}{56} \]
Calculating this gives:
\[ Z \approx \frac{1.09 \times 10^{-5}}{56} \approx 1.94 \approx 2 \]

### Step 8: Determine the Type of Cubic Lattice
From the calculated value of Z:
- Z = 2 corresponds to a Body-Centered Cubic (BCC) structure.

### Conclusion
The type of cubic lattice to which the iron crystal belongs is **Body-Centered Cubic (BCC)**.