Question

An element crystallizes in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g/cm\(^3\). How many atoms are present in 300 g of metal?

• A. \( 12.8 \times 10^{23} \) atoms
• B. \( 6.4 \times 10^{23} \) atoms
• C. \( 3.2 \times 10^{23} \) atoms
• D. \( 1.6 \times 10^{23} \) atoms

Hint:

For bcc crystals, the number of atoms per unit cell is 2. Use the unit cell volume and density to calculate the number of atoms in a given mass.

Answer 1

The Correct Option is B

Step 1: Formula for number of atoms.
The number of atoms can be calculated using the formula: \[ \text{Number of atoms} = \frac{\text{mass of metal}}{\text{molar mass}} \times N_A \] Where: - \( N_A = 6.022 \times 10^{23} \) is Avogadro's number.

Step 2: Calculation.
For a bcc lattice, the number of atoms per unit cell is 2. The volume of the unit cell can be calculated using the formula for the volume of a cube: \[ V_{\text{unit cell}} = a^3 \] Where \( a = 500 \, \text{pm} = 5 \times 10^{-8} \, \text{cm} \), so: \[ V_{\text{unit cell}} = (5 \times 10^{-8})^3 = 1.25 \times 10^{-22} \, \text{cm}^3 \] The density is given as 7.5 g/cm\(^3\), so the mass of the unit cell is: \[ \text{Mass of unit cell} = \text{Density} \times V_{\text{unit cell}} = 7.5 \times 1.25 \times 10^{-22} = 9.375 \times 10^{-22} \, \text{g} \] Now, the number of unit cells in 300 g of the metal is: \[ \frac{300}{9.375 \times 10^{-22}} = 3.2 \times 10^{23} \, \text{unit cells} \] Since each unit cell contains 2 atoms, the total number of atoms is: \[ 3.2 \times 10^{23} \times 2 = 6.4 \times 10^{23} \, \text{atoms} \]
Step 3: Conclusion.
The correct answer is (B) \( 6.4 \times 10^{23} \) atoms.