An element crystallizes in a fcc lattice and the edge length of the unit cell is 0.559 nm . The density of crystal is 3.19 `g` / `cm^3` . Find atomic weight of the element .

Options

A. `100.6`
B. `75.9`
C. `95.8`
D. `83.9`

Answer

To find the atomic weight of the element that crystallizes in a face-centered cubic (fcc) lattice, we can use the following steps:

### Step 1: Determine the number of atoms per unit cell
In a face-centered cubic (fcc) lattice, there are 4 atoms per unit cell. This is because there are 8 corner atoms (each contributing 1/8) and 6 face-centered atoms (each contributing 1/2):
\[ \text{Total atoms} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \text{ atoms} \]

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
where \( a \) is the edge length of the unit cell. Given \( a = 0.559 \, \text{nm} = 0.559 \times 10^{-7} \, \text{cm} \):
\[ V = (0.559 \times 10^{-7} \, \text{cm})^3 = 1.743 \times 10^{-21} \, \text{cm}^3 \]

### Step 3: Use the density to find the mass of the unit cell
The density \( \rho \) is given as \( 3.19 \, \text{g/cm}^3 \). The mass \( m \) of the unit cell can be calculated using the formula:
\[ m = \rho \times V \]
Substituting the values:
\[ m = 3.19 \, \text{g/cm}^3 \times 1.743 \times 10^{-21} \, \text{cm}^3 = 5.56 \times 10^{-21} \, \text{g} \]

### Step 4: Calculate the atomic weight
The atomic weight \( M \) can be calculated using the formula:
\[ M = \frac{m \times N_A}{n} \]
where \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \)) and \( n \) is the number of atoms per unit cell (which is 4 for fcc):
\[ M = \frac{5.56 \times 10^{-21} \, \text{g} \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{4} \]
Calculating this gives:
\[ M = \frac{3.35 \times 10^{3} \, \text{g/mol}}{4} = 838.75 \, \text{g/mol} \]

### Final Answer
The atomic weight of the element is approximately \( 83.88 \, \text{g/mol} \).

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