An element 'X' crystallises as face centred cubic lattice with edge length of 460 pm. The density of the element X, when molar mass of X atom is 60 g `mol^(-1)` is

Options

A. 4.096`" "gcm^(-3)`
B. 2.048`" "gcm^(-3)`
C. 6.144`" "gcm^(-3)`
D. 3.072 g `" "cm^(-3)`

Explanation

To find the density of the element 'X' that crystallizes in a face-centered cubic (FCC) lattice, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters**:
- The edge length (a) of the FCC unit cell is given as 460 pm (picometers).
- The molar mass (M) of the element X is given as 60 g/mol.

2. **Convert edge length from picometers to centimeters**:
\[ a = 460 \, \text{pm} = 460 \times 10^{-10} \, \text{cm} \]

3. **Determine the number of atoms per unit cell (Z)**:
- For a face-centered cubic (FCC) lattice, the number of atoms per unit cell (Z) is 4.

4. **Use the formula for density**:
The formula for density (d) is given by:
\[ d = \frac{Z \times M}{N_A \times a^3} \]
where:
- \(Z\) = number of atoms per unit cell = 4
- \(M\) = molar mass = 60 g/mol
- \(N_A\) = Avogadro's number = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)
- \(a\) = edge length in cm = \(460 \times 10^{-10} \, \text{cm}\)

5. **Calculate \(a^3\)**:
\[ a^3 = (460 \times 10^{-10})^3 = 9.7936 \times 10^{-29} \, \text{cm}^3 \]

6. **Substitute the values into the density formula**:
\[ d = \frac{4 \times 60 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \times 9.7936 \times 10^{-29} \, \text{cm}^3} \]

7. **Calculate the density**:
\[ d = \frac{240 \, \text{g/mol}}{6.022 \times 10^{23} \times 9.7936 \times 10^{-29}} \approx 4.096 \, \text{g/cm}^3 \]

### Final Answer:
The density of element 'X' is approximately **4.096 g/cm³**.