To find the molar mass of the metal with a face-centered cubic (FCC) lattice, we can follow these steps:
### Step 1: Convert the edge length from picometers to centimeters
The given edge length of the unit cell is 404 pm. We need to convert this to centimeters because the density is given in g/cm³.
\[
\text{Edge length (a)} = 404 \, \text{pm} = 404 \times 10^{-12} \, \text{m} = 404 \times 10^{-10} \, \text{cm} = 4.04 \times 10^{-8} \, \text{cm}
\]
### Step 2: Identify the number of atoms in the FCC unit cell
For a face-centered cubic (FCC) lattice, the number of atoms (Z) per unit cell is 4.
\[
Z = 4
\]
### Step 3: Use the density formula to find the molar mass
The formula for density (D) is given by:
\[
D = \frac{Z \cdot M}{a^3 \cdot N_A}
\]
Where:
- \(D\) = density (g/cm³)
- \(M\) = molar mass (g/mol)
- \(a\) = edge length of the unit cell (cm)
- \(N_A\) = Avogadro's number (\(6.02 \times 10^{23} \, \text{mol}^{-1}\))
Rearranging the formula to solve for molar mass (M):
\[
M = \frac{D \cdot a^3 \cdot N_A}{Z}
\]
### Step 4: Substitute the values into the formula
Now we can substitute the known values into the equation:
\[
D = 2.72 \, \text{g/cm}^3
\]
\[
a = 4.04 \times 10^{-8} \, \text{cm}
\]
\[
N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}
\]
\[
Z = 4
\]
Calculating \(a^3\):
\[
a^3 = (4.04 \times 10^{-8})^3 = 6.58 \times 10^{-23} \, \text{cm}^3
\]
Now substituting these values into the molar mass formula:
\[
M = \frac{2.72 \, \text{g/cm}^3 \cdot 6.58 \times 10^{-23} \, \text{cm}^3 \cdot 6.02 \times 10^{23} \, \text{mol}^{-1}}{4}
\]
### Step 5: Calculate the molar mass
Calculating the numerator:
\[
2.72 \cdot 6.58 \times 10^{-23} \cdot 6.02 \times 10^{23} = 2.72 \cdot 39.66 \approx 107.4
\]
Now divide by 4:
\[
M \approx \frac{107.4}{4} \approx 26.85 \, \text{g/mol}
\]
### Final Answer
Thus, the molar mass of the metal is approximately \(26.85 \, \text{g/mol}\), which can be rounded to \(26 \, \text{g/mol}\).
