An elemetnts crystallizes in a face centered cubic lattice and the edge of the unit cell is 0.559nm. The density is `3.19g//cm^(3)`. What is the atomic mass?

Options

A. 87.6
B. 79.9
C. 85.5
D. 83.9

Answer

To find the atomic mass of the element that crystallizes in a face-centered cubic (FCC) lattice, we will use the formula for density:

\[ \text{Density} = \frac{Z \cdot M}{A^3 \cdot N_A} \]

Where:
- \( \text{Density} \) = 3.19 g/cm³
- \( Z \) = number of atoms per unit cell in FCC = 4
- \( M \) = atomic mass (g/mol)
- \( A \) = edge length of the unit cell (cm)
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \) mol⁻¹

### Step 1: Convert the edge length from nanometers to centimeters
Given:
\[ A = 0.559 \text{ nm} = 0.559 \times 10^{-7} \text{ cm} \]

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of the unit cell is given by:
\[ V = A^3 = (0.559 \times 10^{-7} \text{ cm})^3 \]

Calculating this:
\[ V = 0.559^3 \times (10^{-7})^3 = 0.174 \times 10^{-21} \text{ cm}^3 \]

### Step 3: Rearrange the density formula to solve for M
Rearranging the density formula gives:
\[ M = \frac{\text{Density} \cdot A^3 \cdot N_A}{Z} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ M = \frac{3.19 \, \text{g/cm}^3 \cdot 0.174 \times 10^{-21} \, \text{cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{4} \]

### Step 5: Calculate M
Calculating the numerator:
\[ 3.19 \cdot 0.174 \times 10^{-21} \cdot 6.022 \times 10^{23} = 3.19 \cdot 1.048 \approx 3.34 \]

Now divide by 4:
\[ M \approx \frac{3.34}{4} \approx 0.835 \text{ g/mol} \]

### Step 6: Convert to grams per mole
Since we need the atomic mass in grams per mole, we multiply by 1000:
\[ M \approx 83.5 \text{ g/mol} \]

Thus, the atomic mass of the element is approximately **83.5 g/mol**.

### Final Answer
The atomic mass of the element is approximately **83.5 g/mol**.

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