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 the element crystallizing in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Understand the BCC Structure
In a body-centered cubic lattice, there are 2 atoms per unit cell (Z = 2).

### Step 2: Convert Edge Length to Centimeters
The edge length (a) is given as 500 pm (picometers). We need to convert this to centimeters:
\[ 1 \text{ pm} = 10^{-12} \text{ m} = 10^{-10} \text{ cm} \]
So,
\[ a = 500 \text{ pm} = 500 \times 10^{-10} \text{ cm} = 5 \times 10^{-8} \text{ cm} \]

### Step 3: Use the Density Formula
The formula for density (D) in terms of the number of atoms per unit cell (Z), molar mass (M), edge length (a), and Avogadro's number (N_A) is:
\[ D = \frac{Z \cdot M}{a^3 \cdot N_A} \]
Where:
- D = density (4 g/cm³)
- Z = number of atoms per unit cell (2 for BCC)
- M = molar mass (what we are trying to find)
- a = edge length in cm (5 × 10⁻⁸ cm)
- N_A = Avogadro's number (6 × 10²³ mol⁻¹)

### Step 4: Rearrange the Formula to Solve for M
Rearranging the density formula gives us:
\[ M = \frac{D \cdot a^3 \cdot N_A}{Z} \]

### Step 5: Substitute the Values
Now, substituting the known values into the equation:
\[ M = \frac{4 \, \text{g/cm}^3 \cdot (5 \times 10^{-8} \, \text{cm})^3 \cdot (6 \times 10^{23} \, \text{mol}^{-1})}{2} \]

### Step 6: Calculate \( a^3 \)
Calculating \( a^3 \):
\[ (5 \times 10^{-8})^3 = 125 \times 10^{-24} \text{ cm}^3 = 1.25 \times 10^{-22} \text{ cm}^3 \]

### Step 7: Substitute \( a^3 \) Back into the Equation
Now substituting \( a^3 \) back into the equation:
\[ M = \frac{4 \cdot 1.25 \times 10^{-22} \cdot 6 \times 10^{23}}{2} \]

### Step 8: Calculate the Molar Mass
Calculating the above expression:
\[ M = \frac{4 \cdot 1.25 \cdot 6}{2} = \frac{30}{2} = 15 \, \text{g/mol} \]
However, we need to multiply by \( 10^{-1} \) to adjust for the units:
\[ M = 150 \, \text{g/mol} \]

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

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