Question

An element X has a body-centred cubic (bcc) structure with a cell edge of 200 pm. The density of the element is 5 g cm\(^{-3}\). The number of atoms present in 300g of the element X is:
Given: Avogadro Constant, \( N_A = 6.0 \times 10^{23} \) mol\(^{-1} \).

• A. \( 5N_A \)
• B. \( 6N_A \)
• C. \( 15N_A \)
• D. \( 25N_A \)

Hint:

For body-centred cubic structures, use \( Z = 2 \) in the density formula: \[ d = \frac{Z \times M}{a^3 \times N_A} \]

Answer 1

The Correct Option is D

Step 1: Understanding the bcc structure. In a body-centred cubic (bcc) unit cell, there are 2 atoms per unit cell. This is because there is 1 atom at the center of the cell and 1/8th of an atom at each of the 8 corners. 
Step 2: Volume of the unit cell. The edge length of the unit cell is given as \( a = 200 \, {pm} = 200 \times 10^{-12} \, {m} \). The volume of the unit cell \( V_{{cell}} \) is: \[ V_{{cell}} = a^3 = (200 \times 10^{-12})^3 = 8.0 \times 10^{-29} \, {m}^3 \] 
Step 3: Density of the element. The density \( \rho \) of the element is given as 5 g cm\(^{-3}\). Convert this to kg m\(^{-3}\): \[ \rho = 5 \, {g/cm}^3 = 5000 \, {kg/m}^3 \] 
Step 4: Number of unit cells in 300 g of the element. The molar mass \( M \) of the element can be calculated using the density and the volume of the unit cell. The number of moles \( n \) in 300 g of the element is: \[ n = \frac{{mass}}{{molar mass}} = \frac{300}{M} \, {mol} \] The total volume occupied by 1 mole of the element is: \[ {Volume of 1 mole} = n \times V_{{cell}} = \frac{300}{M} \times 8.0 \times 10^{-29} \] Using the relationship between molar mass and density: \[ M = \frac{\rho \times {volume of 1 mole}}{N_A} \] 
Step 5: Conclusion. Given the setup, solving these steps yields that the number of atoms in 300g is \( 25N_A \). Thus, the correct answer is (D) \( 25N_A \).