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 solve the problem, we need to calculate the density of the element that crystallizes in a face-centered cubic (FCC) structure. Here are the step-by-step calculations:

### Step 1: Convert the edge length from picometers to centimeters
Given edge length \( a = 200 \) pm (picometers).

1 pm = \( 10^{-10} \) cm, therefore:
\[ a = 200 \, \text{pm} = 200 \times 10^{-10} \, \text{cm} = 2 \times 10^{-8} \, \text{cm} \]

### Step 2: Determine the number of atoms in 400 g of the element
We are given that 400 g of the element contains \( 48 \times 10^{23} \) atoms.

### Step 3: Calculate the mass of one atom
To find the mass of one atom, we can use the formula:
\[ \text{Mass of one atom} = \frac{\text{Total mass}}{\text{Number of atoms}} = \frac{400 \, \text{g}}{48 \times 10^{23}} \]
Calculating this gives:
\[ \text{Mass of one atom} = \frac{400}{48 \times 10^{23}} = \frac{400}{48} \times 10^{-23} \approx 8.33 \times 10^{-23} \, \text{g} \]

### Step 4: Calculate the molar mass (M) of the element
Using Avogadro's number \( N_A = 6.023 \times 10^{23} \):
\[ M = \text{Mass of one atom} \times N_A = 8.33 \times 10^{-23} \, \text{g} \times 6.023 \times 10^{23} \approx 50.18 \, \text{g/mol} \]

### Step 5: Determine the number of atoms per unit cell (Z) for FCC
In a face-centered cubic (FCC) structure, the number of atoms per unit cell \( Z = 4 \).

### Step 6: Calculate the volume of the unit cell
The volume \( V \) of the cubic 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 7: Calculate the density (D)
The density can be calculated using the formula:
\[ D = \frac{Z \times M}{V \times N_A} \]
Substituting the values:
\[ D = \frac{4 \times 50.18 \, \text{g/mol}}{8 \times 10^{-24} \, \text{cm}^3 \times 6.023 \times 10^{23}} \]
Calculating this gives:
\[ D = \frac{200.72}{4.8184 \times 10^{-1}} \approx 41.656 \, \text{g/cm}^3 \]

### Final Answer
The density of the element is approximately \( 41.656 \, \text{g/cm}^3 \).

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