An element crystallising in body centred cublic lattice has edge length of 500 pm. If the density is 4 g `cm^(-3)`, the atomic mass of the element `("in g mol"^(-1))` is (consider `N_(A)=6xx10^(23))`

Options

A. 100
B. 250
C. 125
D. 150

Explanation

To find the atomic mass of an element crystallizing in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Identify the parameters given in the problem
- Edge length (A) = 500 pm
- Density (ρ) = 4 g/cm³
- Avogadro's number (N_A) = 6 x 10²³

### Step 2: Convert the edge length from picometers to centimeters
Since the density is given in g/cm³, we need to convert the edge length from picometers to centimeters:
\[ A = 500 \text{ pm} = 500 \times 10^{-10} \text{ cm} \]

### Step 3: Calculate the volume of the unit cell
The volume (V) of the unit cell can be calculated using the formula:
\[ V = A^3 \]
Substituting the value of A:
\[ V = (500 \times 10^{-10} \text{ cm})^3 \]
\[ V = 1.25 \times 10^{-22} \text{ cm}^3 \]

### Step 4: Determine the number of atoms in the BCC unit cell
In a body-centered cubic lattice, there are 2 atoms per unit cell (Z = 2).

### Step 5: Use the formula for density to find the molar mass (m)
The formula for density is given by:
\[ \rho = \frac{Z \cdot m}{N_A \cdot V} \]
Rearranging the formula to solve for m:
\[ m = \frac{\rho \cdot N_A \cdot V}{Z} \]
Substituting the known values:
- ρ = 4 g/cm³
- N_A = 6 x 10²³
- V = 1.25 x 10^{-22} cm³
- Z = 2

### Step 6: Calculate the atomic mass (m)
Substituting the values into the equation:
\[ m = \frac{4 \text{ g/cm}^3 \cdot 6 \times 10^{23} \cdot 1.25 \times 10^{-22} \text{ cm}^3}{2} \]
Calculating:
\[ m = \frac{4 \cdot 6 \cdot 1.25}{2} \]
\[ m = \frac{30}{2} = 15 \text{ g/mol} \]

### Step 7: Final calculation
Now, we need to multiply by 10 to convert to grams per mole:
\[ m = 150 \text{ g/mol} \]

### Conclusion
The atomic mass of the element is **150 g/mol**.

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