An element has atomic mass `93 g mol^(-1)` and density `11.5 g cm^(-3)`. If the edge length of its unit cell is 300 pm, identify the type of unit cell.

Explanation

To determine the type of unit cell for the given element with an atomic mass of 93 g/mol, a density of 11.5 g/cm³, and an edge length of 300 pm, we can follow these steps:

### Step 1: Convert Edge Length to Centimeters
The edge length is given in picometers (pm). We need to convert it to centimeters (cm) for consistency with the density units.

\[ \text{Edge length (A)} = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m} = 300 \times 10^{-10} \, \text{cm} = 3.00 \times 10^{-8} \, \text{cm} \]

### Step 2: Use the Density Formula
The formula for density (d) in terms of the number of atoms per unit cell (Z), molar mass (M), and edge length (a) is given by:

\[ d = \frac{Z \cdot M}{A^3 \cdot N_0} \]

Where:
- \(d\) = density (g/cm³)
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass (g/mol)
- \(A\) = edge length (cm)
- \(N_0\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))

### Step 3: Rearrange the Formula to Solve for Z
Rearranging the density formula to solve for Z gives:

\[ Z = \frac{d \cdot A^3 \cdot N_0}{M} \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:

- \(d = 11.5 \, \text{g/cm}^3\)
- \(A = 3.00 \times 10^{-8} \, \text{cm}\)
- \(M = 93 \, \text{g/mol}\)
- \(N_0 = 6.022 \times 10^{23} \, \text{mol}^{-1}\)

Calculating \(A^3\):

\[ A^3 = (3.00 \times 10^{-8})^3 = 2.7 \times 10^{-24} \, \text{cm}^3 \]

Now substituting these values into the equation for Z:

\[ Z = \frac{11.5 \cdot (2.7 \times 10^{-24}) \cdot (6.022 \times 10^{23})}{93} \]

### Step 5: Calculate Z
Calculating the numerator:

\[ 11.5 \cdot 2.7 \times 10^{-24} \cdot 6.022 \times 10^{23} = 11.5 \cdot 2.7 \cdot 6.022 \times 10^{-1} \approx 11.5 \cdot 16.27 \approx 187.1 \]

Now divide by 93:

\[ Z = \frac{187.1}{93} \approx 2.01 \]

### Step 6: Identify the Type of Unit Cell
Since Z is approximately 2, we can identify the type of unit cell. A unit cell with Z = 2 corresponds to a Body-Centered Cubic (BCC) structure.

### Final Answer
The type of unit cell for the given element is **Body-Centered Cubic (BCC)**.

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