Titanium metal has a density of `4.54 "g cm"^(-3)` and an edge length of 412.6 pm. In what cubic unit cell does titanium crystallise ?

Explanation

To determine the type of cubic unit cell in which titanium crystallizes, we will follow these steps:

### Step 1: Identify the given data
- Density of titanium (D) = 4.54 g/cm³
- Edge length (a) = 412.6 pm = 412.6 x 10⁻¹² m = 4.126 x 10⁻⁸ cm
- Atomic mass of titanium (m) = 48 g/mol
- Avogadro's number (Nₐ) = 6.022 x 10²³ mol⁻¹

### Step 2: Use the formula for density in terms of unit cell parameters
The formula for density (D) in terms of the number of atoms per unit cell (Z), atomic mass (m), and the volume of the unit cell (a³) is given by:

\[ D = \frac{Z \times m}{N_a \times a^3} \]

### Step 3: Rearrange the formula to solve for Z
We can rearrange the formula to find Z:

\[ Z = \frac{D \times N_a \times a^3}{m} \]

### Step 4: Substitute the values into the equation
Now we will substitute the known values into the equation:

1. Convert edge length from pm to cm:
\[ a = 412.6 \, \text{pm} = 4.126 \times 10^{-8} \, \text{cm} \]

2. Calculate \(a^3\):
\[ a^3 = (4.126 \times 10^{-8})^3 = 6.98 \times 10^{-24} \, \text{cm}^3 \]

3. Substitute the values into the equation for Z:
\[ Z = \frac{4.54 \, \text{g/cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1} \times 6.98 \times 10^{-24} \, \text{cm}^3}{48 \, \text{g/mol}} \]

### Step 5: Perform the calculations
Calculating the numerator:
\[ 4.54 \times 6.022 \times 6.98 \approx 189.34 \]

Now divide by the atomic mass:
\[ Z \approx \frac{189.34}{48} \approx 3.95 \approx 4 \]

### Step 6: Determine the type of unit cell
Since Z = 4, titanium crystallizes in a face-centered cubic (FCC) unit cell.

### Final Answer
Titanium crystallizes in a face-centered cubic (FCC) unit cell.

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