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 can use the following formula related to density:

\[ \text{Density} = \frac{Z \times M}{V} \]

Where:
- \( Z \) = number of atoms per unit cell (for FCC, \( Z = 4 \))
- \( M \) = atomic mass (in grams per mole)
- \( V \) = volume of the unit cell (in cm³)

### Step-by-Step Solution:

1. **Convert the edge length to centimeters**:
The edge length \( a \) is given as \( 0.559 \, \text{nm} \). To convert nanometers to centimeters, we use the conversion factor \( 1 \, \text{nm} = 10^{-7} \, \text{cm} \):
\[ a = 0.559 \, \text{nm} = 0.559 \times 10^{-7} \, \text{cm} = 5.59 \times 10^{-8} \, \text{cm} \]

2. **Calculate the volume of the unit cell**:
The volume \( V \) of the unit cell can be calculated using the formula for the volume of a cube:
\[ V = a^3 = (5.59 \times 10^{-8} \, \text{cm})^3 \]
\[ V = 1.75 \times 10^{-22} \, \text{cm}^3 \]

3. **Use the density to find the atomic mass**:
Rearranging the density formula gives:
\[ M = \text{Density} \times V / Z \]
Given that the density is \( 3.19 \, \text{g/cm}^3 \) and \( Z = 4 \) for FCC:
\[ M = \frac{3.19 \, \text{g/cm}^3 \times 1.75 \times 10^{-22} \, \text{cm}^3}{4} \]
\[ M = \frac{5.57 \times 10^{-22} \, \text{g}}{4} = 1.3925 \times 10^{-22} \, \text{g} \]

4. **Convert the mass to atomic mass units**:
Since \( 1 \, \text{g/mol} = 6.022 \times 10^{23} \, \text{atoms/mol} \), we convert the mass to atomic mass:
\[ M = 1.3925 \times 10^{-22} \, \text{g} \times 6.022 \times 10^{23} \, \text{atoms/mol} \]
\[ M \approx 83.78 \, \text{g/mol} \]

### Final Answer:
The atomic mass of the element is approximately \( 83.78 \, \text{g/mol} \).