The density of argon (face centered cubic cell) is `1.83g//cm^(3) at 20^(@)C`. What is the length of an edge a unit cell? `("Atomic mass": Ar=40)`

Options

A. 0.599nm
B. 0.569nm
C. 0.525nm
D. 0.551nm

Answer

To find the length of an edge \( a \) of the unit cell of argon in a face-centered cubic (FCC) structure, we can use the relationship between density, molar mass, and the volume of the unit cell. Here’s a step-by-step solution:

### Step 1: Write down the formula for density
The density \( \rho \) of a substance can be expressed as:
\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]
where:
- \( Z \) = number of atoms per unit cell (for FCC, \( Z = 4 \))
- \( M \) = molar mass of the substance (for argon, \( M = 40 \, \text{g/mol} \))
- \( N_A \) = Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \))
- \( a \) = length of the edge of the unit cell (in cm)

### Step 2: Rearrange the formula to solve for \( a^3 \)
Rearranging the formula gives:
\[ a^3 = \frac{Z \cdot M}{\rho \cdot N_A} \]

### Step 3: Substitute the known values into the equation
Given:
- \( \rho = 1.83 \, \text{g/cm}^3 \)
- \( Z = 4 \)
- \( M = 40 \, \text{g/mol} \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Substituting these values into the equation:
\[ a^3 = \frac{4 \cdot 40}{1.83 \cdot 6.022 \times 10^{23}} \]

### Step 4: Calculate \( a^3 \)
Calculating the numerator:
\[ 4 \cdot 40 = 160 \]

Calculating the denominator:
\[ 1.83 \cdot 6.022 \times 10^{23} \approx 1.101 \times 10^{24} \]

Now substituting these into the equation:
\[ a^3 = \frac{160}{1.101 \times 10^{24}} \approx 1.45 \times 10^{-23} \, \text{cm}^3 \]

### Step 5: Calculate \( a \)
Now, take the cube root to find \( a \):
\[ a = (1.45 \times 10^{-23})^{1/3} \approx 5.25 \times 10^{-8} \, \text{cm} \]

### Step 6: Convert \( a \) to nanometers
To convert centimeters to nanometers:
\[ a = 5.25 \times 10^{-8} \, \text{cm} \times 10^7 \, \text{nm/cm} = 5.25 \times 10^{-1} \, \text{nm} = 5.25 \, \text{nm} \]

### Final Answer
The length of an edge \( a \) of the unit cell is approximately:
\[ a \approx 5.25 \, \text{Å} \text{ or } 0.525 \, \text{nm} \]