An element (atomic mass `= 100 g//mol`) having bcc structure has unit cell edge 400 pm .Them density of the element is

Options

A. `10.38gm//cm^(3)`
B. `5.19gm//cm^(3)`
C. `7.29gm//cm^(3)`
D. `2.14gm//cm^(3)`

Answer

To find the density of the element with a body-centered cubic (BCC) structure, we can follow these steps:

### Step 1: Identify the given data
- Atomic mass (M) = 100 g/mol
- Unit cell edge length (a) = 400 pm = 400 × 10^(-10) cm
- For BCC structure, the number of atoms per unit cell (Z) = 2

### Step 2: Convert the edge length to centimeters
The edge length is given in picometers (pm). We need to convert it to centimeters (cm):
\[ a = 400 \, \text{pm} = 400 \times 10^{-10} \, \text{cm} \]

### Step 3: 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:
\[ V = (400 \times 10^{-10} \, \text{cm})^3 \]
\[ V = 64 \times 10^{-30} \, \text{cm}^3 = 6.4 \times 10^{-29} \, \text{cm}^3 \]

### Step 4: Use the formula for density
The density (ρ) of the element can be calculated using the formula:
\[ \rho = \frac{Z \cdot M}{N_A \cdot V} \]
Where:
- \( Z \) = number of atoms per unit cell = 2
- \( M \) = molar mass = 100 g/mol
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( V \) = volume of the unit cell

### Step 5: Substitute the values into the density formula
Now substituting the values we have:
\[ \rho = \frac{2 \cdot 100 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 6.4 \times 10^{-29} \, \text{cm}^3} \]

### Step 6: Calculate the density
Calculating the numerator:
\[ 2 \cdot 100 = 200 \, \text{g} \]
Calculating the denominator:
\[ 6.022 \times 10^{23} \cdot 6.4 \times 10^{-29} \approx 3.85 \times 10^{-5} \, \text{g/cm}^3 \]
Now, substituting back into the density formula:
\[ \rho = \frac{200}{3.85 \times 10^{-5}} \approx 5.19 \, \text{g/cm}^3 \]

### Final Answer
The density of the element is approximately **5.19 g/cm³**.

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