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, we can follow these steps:

### Step 1: Understand the Given Data
- Specific Gravity (SG) = 10.2
- Density (ρ) = 10.2 g/cm³ (since SG is the ratio of the density of a substance to the density of water)
- Unit cell type = Body-Centered Cubic (BCC)
- Unit cell edge length (a) = 3.147 Å = 3.147 × 10⁻⁸ cm (conversion from angstroms to centimeters)

### Step 2: Calculate the Volume of the Unit Cell
The volume (V) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (3.147 \times 10^{-8} \, \text{cm})^3 \]

### Step 3: Calculate the Volume
Calculating the volume:
\[ V = (3.147 \times 10^{-8})^3 = 3.116 \times 10^{-24} \, \text{cm}^3 \]

### Step 4: Determine the Number of Atoms per Unit Cell (Z)
For a body-centered cubic (BCC) structure, the number of atoms per unit cell (Z) is 2.

### Step 5: Use the Density Formula
The density (ρ) of the metal can be expressed using the formula:
\[ \rho = \frac{Z \times \text{Atomic Weight}}{V \times N_A} \]
Where:
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Rearranging the formula to solve for Atomic Weight (A):
\[ A = \frac{\rho \times V \times N_A}{Z} \]

### Step 6: Substitute the Known Values
Substituting the known values into the equation:
\[ A = \frac{10.2 \, \text{g/cm}^3 \times 3.116 \times 10^{-24} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{2} \]

### Step 7: Calculate the Atomic Weight
Calculating the numerator:
\[ 10.2 \times 3.116 \times 10^{-24} \times 6.022 \times 10^{23} = 10.2 \times 3.116 \times 6.022 \times 10^{-1} \]
\[ = 10.2 \times 18.74 \approx 191.1 \]

Now divide by 2:
\[ A = \frac{191.1}{2} \approx 95.55 \, \text{g/mol} \]

### Final Result
The atomic weight of the unknown metal is approximately **95.55 g/mol**.

---