An element crystallises in a structure having a fcc unit cell of an edge 200 pm. Calculate its density if 400 g of this element contain `48xx10^(23)` atoms.

Explanation

To calculate the density of the element that crystallizes in a face-centered cubic (FCC) structure with an edge length of 200 pm, we can follow these steps:

### Step 1: Identify the parameters
- **Edge length (A)** = 200 pm = \(200 \times 10^{-12}\) m
- **Number of atoms (N)** = \(48 \times 10^{23}\) atoms
- **Mass (m)** = 400 g = 0.4 kg (since we will use SI units)
- **Avogadro's number (N_A)** = \(6.022 \times 10^{23}\) atoms/mol
- **Number of atoms per unit cell (Z)** for FCC = 4

### Step 2: Calculate the molecular mass (M)
The molecular mass can be calculated using the formula:
\[ M = \frac{\text{mass of the sample}}{\text{number of moles}} \]
First, we need to calculate the number of moles:
\[ \text{Number of moles} = \frac{N}{N_A} = \frac{48 \times 10^{23}}{6.022 \times 10^{23}} \approx 7.97 \text{ moles} \]
Now, we can calculate the molecular mass:
\[ M = \frac{0.4 \text{ kg}}{7.97 \text{ moles}} \approx 0.0503 \text{ kg/mol} = 50.3 \text{ g/mol} \]

### Step 3: Convert edge length to cm
We need to convert the edge length from pm to cm for the density calculation:
\[ A = 200 \text{ pm} = 200 \times 10^{-12} \text{ m} = 200 \times 10^{-10} \text{ cm} = 2 \times 10^{-8} \text{ cm} \]

### Step 4: Calculate the volume of the unit cell (V)
The volume of the unit cell can be calculated using the formula:
\[ V = A^3 = (2 \times 10^{-8} \text{ cm})^3 = 8 \times 10^{-24} \text{ cm}^3 \]

### Step 5: Calculate the density (D)
The density can be calculated using the formula:
\[ D = \frac{Z \times M}{V \times N_A} \]
Substituting the values:
\[ D = \frac{4 \times 50.3 \text{ g/mol}}{8 \times 10^{-24} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \]
Calculating the denominator:
\[ 8 \times 10^{-24} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1} \approx 4.82 \times 10^{-1} \text{ cm}^3 \]
Now substituting back into the density equation:
\[ D = \frac{4 \times 50.3}{4.82 \times 10^{-1}} \approx \frac{201.2}{0.482} \approx 417.4 \text{ g/cm}^3 \]

### Final Answer
The density of the element is approximately **417.4 g/cm³**.

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