An unknown metal is found to have a specific gravity of 10.2 at `25^(@)C`. It is found to crystallize in a body centered cubic lattice with a unit cell edge length of `3.147 Å`. Calculate the atomic weight.

Answer

To calculate the atomic weight of the unknown metal based on the given information, we will follow these steps:

### Step 1: Understand the relationship between specific gravity and density
Specific gravity (SG) is defined as the ratio of the density of a substance to the density of a reference substance (usually water). The formula is:
\[ \text{SG} = \frac{\text{Density of metal}}{\text{Density of water}} \]

Given:
- Specific gravity (SG) = 10.2
- Density of water at 25°C ≈ 0.997 g/cm³

### Step 2: Calculate the density of the metal
Using the specific gravity, we can calculate the density of the metal (D):
\[ D = \text{SG} \times \text{Density of water} \]
\[ D = 10.2 \times 0.997 \, \text{g/cm}^3 \]
\[ D \approx 10.1694 \, \text{g/cm}^3 \]

### Step 3: Convert the density to kg/m³
To use the density in the formula for calculating atomic weight, we need to convert it to kg/m³:
\[ D = 10.1694 \, \text{g/cm}^3 = 10.1694 \times 1000 \, \text{kg/m}^3 = 10169.4 \, \text{kg/m}^3 \]

### Step 4: Identify the parameters for the BCC lattice
In a body-centered cubic (BCC) lattice:
- Z (number of atoms per unit cell) = 2 (1 atom from the center and 1/8 from each of the 8 corners)
- Edge length (a) = 3.147 Å = \( 3.147 \times 10^{-10} \, \text{m} \)

### Step 5: Use the formula for density in terms of atomic weight
The formula for density (D) in a crystal lattice is given by:
\[ D = \frac{Z \cdot M}{a^3 \cdot N_0} \]
Where:
- D = density (kg/m³)
- Z = number of atoms per unit cell
- M = atomic weight (kg/mol)
- a = edge length (m)
- \( N_0 \) = Avogadro's number \( \approx 6.022 \times 10^{23} \, \text{mol}^{-1} \)

### Step 6: Rearrange the formula to solve for atomic weight (M)
Rearranging the formula gives:
\[ M = \frac{D \cdot a^3 \cdot N_0}{Z} \]

### Step 7: Substitute the known values into the formula
Substituting the values we have:
- D = 10169.4 kg/m³
- a = \( 3.147 \times 10^{-10} \, \text{m} \)
- \( N_0 = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- Z = 2

Calculating \( a^3 \):
\[ a^3 = (3.147 \times 10^{-10})^3 = 3.11 \times 10^{-29} \, \text{m}^3 \]

Now substituting into the formula:
\[ M = \frac{10169.4 \times 3.11 \times 10^{-29} \times 6.022 \times 10^{23}}{2} \]

### Step 8: Perform the calculation
Calculating the numerator:
\[ 10169.4 \times 3.11 \times 10^{-29} \times 6.022 \times 10^{23} \approx 1.93 \times 10^{-5} \]

Now divide by 2:
\[ M \approx \frac{1.93 \times 10^{-5}}{2} \approx 9.65 \times 10^{-6} \, \text{kg/mol} \]

### Step 9: Convert kg/mol to g/mol
\[ M \approx 9.65 \times 10^{-6} \times 1000 \approx 96.5 \, \text{g/mol} \]

### Final Answer
The atomic weight of the unknown metal is approximately **96.5 g/mol**.

---