A element cystallizes in fcc lattice and edge length of unit cell is 400 pm. If density of unit cell is `11.2 g cm^(-3)`, then atomic mass of the element is

Options

A. 215.6
B. 431.2
C. 107.8
D. 98.6

Explanation

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

### Step 1: Identify the parameters
- **Lattice type**: FCC
- **Edge length (a)**: 400 pm = 400 x 10^-10 cm
- **Density (d)**: 11.2 g/cm³
- **Number of atoms per unit cell (Z)**: For FCC, Z = 4

### Step 2: Convert edge length to cm
Given:
- Edge length \( a = 400 \, \text{pm} = 400 \times 10^{-10} \, \text{cm} = 4 \times 10^{-8} \, \text{cm} \)

### Step 3: Calculate the volume of the unit cell
The volume \( V \) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of \( a \):
\[ V = (4 \times 10^{-8} \, \text{cm})^3 = 64 \times 10^{-24} \, \text{cm}^3 = 6.4 \times 10^{-23} \, \text{cm}^3 \]

### Step 4: Use the density formula to find the atomic mass
The formula for density \( d \) is given by:
\[ d = \frac{Z \cdot M}{V \cdot N_A} \]
Where:
- \( M \) = atomic mass (g/mol)
- \( N_A \) = Avogadro's number \( \approx 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Rearranging the formula to solve for \( M \):
\[ M = \frac{d \cdot V \cdot N_A}{Z} \]

### Step 5: Substitute the known values
Substituting the values into the equation:
\[ M = \frac{11.2 \, \text{g/cm}^3 \cdot 6.4 \times 10^{-23} \, \text{cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{4} \]

### Step 6: Calculate the atomic mass
Calculating the numerator:
\[ 11.2 \cdot 6.4 \cdot 6.022 \approx 11.2 \cdot 38.4648 \approx 430.6 \]
Now divide by 4:
\[ M \approx \frac{430.6}{4} \approx 107.65 \, \text{g/mol} \]

### Conclusion
The atomic mass of the element is approximately **107.65 g/mol**.

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