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

Explanation:
To determine the density of the element with the given properties, follow these steps:

1. Understand the Given Information:
– Atomic Mass (A): 100 g/mol
– Unit Cell Structure: Body-Centered Cubic (bcc)
– Edge Length of Unit Cell (a): 400 picometers (pm)

2. Convert Edge Length to Centimeters:
– 1 pm = 1 × 10^-12 meters
– 400 pm = 400 × 10^-12 meters = 4 × 10^-10 meters
– Convert meters to centimeters: 4 × 10^-10 meters = 4 × 10^-8 centimeters

3. Calculate Volume of the Unit Cell:
– Volume (V) = a³
– V = (4 × 10^-8 cm)³ = 64 × 10^-24 cm³

4. Determine Number of Atoms per Unit Cell in bcc Structure:
– In a bcc lattice, there are 2 atoms per unit cell.

5. Calculate the Mass of Atoms in the Unit Cell:
– Molar mass (A) = 100 g/mol
– Avogadro’s number (Nₐ) = 6.022 × 10^23 atoms/mol
– Mass of one atom = A / Nₐ = 100 / 6.022 × 10^23 ≈ 1.66 × 10^-22 g
– Total mass in unit cell = 2 atoms × 1.66 × 10^-22 g ≈ 3.32 × 10^-22 g

6. Compute Density (ρ):
– Density formula: ρ = mass/volume
– ρ = 3.32 × 10^-22 g / 64 × 10^-24 cm³ ≈ 5.188 g/cm³

Therefore, the density of the element is approximately 5.188 g/cm³, which corresponds to Option B.