To solve the problem, we need to find the edge length of the unit cell of NaCl and the volume of the cube containing one mole of NaCl using the given density.
### Step 1: Understand the formula for density
The density (\( \rho \)) of a crystalline solid can be expressed using the formula:
\[
\rho = \frac{n \cdot M}{V \cdot N_A}
\]
where:
- \( n \) = number of formula units per unit cell
- \( M \) = molar mass of the compound
- \( V \) = volume of the unit cell
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \))
### Step 2: Identify the values
For NaCl:
- The molar mass (\( M \)) of NaCl = 58.5 g/mol
- The density (\( \rho \)) = 2.165 g/cm³
- For NaCl, which has a face-centered cubic (FCC) structure, \( n = 4 \).
### Step 3: Calculate the volume of the unit cell
Rearranging the density formula to find the volume of the unit cell (\( V \)):
\[
V = \frac{n \cdot M}{\rho \cdot N_A}
\]
Substituting the known values:
\[
V = \frac{4 \cdot 58.5 \, \text{g/mol}}{2.165 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}
\]
### Step 4: Perform the calculation
Calculating the volume:
\[
V = \frac{234 \, \text{g/mol}}{2.165 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 1.794 \times 10^{-22} \, \text{cm}^3
\]
### Step 5: Find the edge length of the unit cell
The volume of the unit cell (\( V \)) is also related to the edge length (\( a \)) of the cube:
\[
V = a^3
\]
Thus,
\[
a = \sqrt[3]{V}
\]
Substituting the volume we calculated:
\[
a = \sqrt[3]{1.794 \times 10^{-22} \, \text{cm}^3} \approx 5.64 \times 10^{-8} \, \text{cm}
\]
### Step 6: Calculate the volume of the cube containing one mole of NaCl
The molar volume (\( V_m \)) can be calculated using:
\[
V_m = \frac{M}{\rho}
\]
Substituting the values:
\[
V_m = \frac{58.5 \, \text{g/mol}}{2.165 \, \text{g/cm}^3} \approx 27.0 \, \text{cm}^3
\]
### Final Answers
- The edge length of the unit cell of NaCl is approximately \( 5.64 \times 10^{-8} \, \text{cm} \).
- The volume of the cube containing one mole of NaCl is approximately \( 27.0 \, \text{cm}^3 \).
