A unit of cell of sodium chloride has four formula units. The edge length of the unit cell is `0.564 nm`. What is the density of sodium chloride?

Options

A. `"3.89 g cm"^(-3)`
B. `"2.16 g cm"^(-3)`
C. `"3 g cm"^(-3)`
D. `"1.82 g cm"^(-3)`

Explanation

To find the density of sodium chloride (NaCl) given that the unit cell has four formula units and an edge length of 0.564 nm, we can follow these steps:

### Step 1: Understand the formula for density
The density (\( \rho \)) of a crystal can be calculated using the formula:
\[ \rho = \frac{Z \cdot M}{A^3 \cdot N_A} \]
where:
- \( Z \) = number of formula units per unit cell
- \( M \) = molar mass of the substance (in grams per mole)
- \( A \) = edge length of the unit cell (in centimeters)
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol\(^{-1}\))

### Step 2: Identify the values
From the problem:
- \( Z = 4 \) (since there are four formula units of NaCl per unit cell)
- The molar mass of NaCl = 58.5 g/mol
- The edge length \( A = 0.564 \) nm = \( 0.564 \times 10^{-7} \) cm

### Step 3: Calculate \( A^3 \)
First, we need to calculate \( A^3 \):
\[ A^3 = (0.564 \times 10^{-7} \text{ cm})^3 = 0.564^3 \times 10^{-21} \text{ cm}^3 \]
Calculating \( 0.564^3 \):
\[ 0.564^3 \approx 0.179 \]
Thus,
\[ A^3 \approx 0.179 \times 10^{-21} \text{ cm}^3 \]

### Step 4: Substitute values into the density formula
Now we can substitute the values into the density formula:
\[ \rho = \frac{4 \cdot 58.5}{0.179 \times 10^{-21} \cdot 6.022 \times 10^{23}} \]
Calculating the denominator:
\[ 0.179 \times 10^{-21} \cdot 6.022 \times 10^{23} \approx 1.077 \times 10^{3} \]
Now substituting this back into the density formula:
\[ \rho = \frac{234}{1.077 \times 10^{3}} \approx 0.217 \text{ g/cm}^3 \]

### Step 5: Final calculation
Now, calculating the density:
\[ \rho \approx 2.16 \text{ g/cm}^3 \]

### Conclusion
Thus, the density of sodium chloride is approximately \( 2.16 \text{ g/cm}^3 \).