Question

A diatomic molecule $X_{2}$ has a body-centred cubic (bcc) structure with a cell edge of $300\, pm$. The density of the molecule is $6.17 \,g \,cm ^{-3}$. The number of molecules present in $200\, g$ of $X _{2}$ is (Avogadro constant $\left.\left( N _{ A }\right)=6 \times 10^{23} \,mol ^{-1}\right)$

• A. $8\, N _{ A }$
• B. $40\, N _{ A }$
• C. $4 \,N _{ A }$
• D. $2 \,N _{ A }$

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the number of molecules present in \(200 \, \text{g}\) of \(X_2\) given its density and the dimensions of its body-centered cubic structure. Here’s the step-by-step calculation:

1. Calculate the volume of one unit cell:
   • Given that the cell edge length is \(300 \, \text{pm}\), we first convert this into centimeters: \( 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m} = 3 \times 10^{-8} \, \text{cm} \).
   • The volume of the cubic unit cell is: \( V = a^3 = (3 \times 10^{-8} \, \text{cm})^3 = 27 \times 10^{-24} \, \text{cm}^3 \).
2. Determine the number of molecules per unit cell:
   • In a body-centered cubic (bcc) structure, there are two atoms per unit cell. Since the unit cell contains \(X_2\) molecules, one molecule comprises two atoms.
   • Thus, each bcc unit cell contains one molecule of \(X_2\).
3. Calculate the mass of one unit cell:
   • The given density of the molecule is \( 6.17 \, \text{g/cm}^3 \).
   • Using the formula: \( \text{Mass} = \text{Density} \times \text{Volume} \), we get: \( \text{Mass of one unit cell} = 6.17 \times 27 \times 10^{-24} \, \text{g} \).
   • Thus, \( \text{Mass of one unit cell} = 166.59 \times 10^{-24} \, \text{g} \).
4. Find the molar mass of \(X_2\):
   • Since one unit cell contains one molecule of \(X_2\) and its mass is \(166.59 \times 10^{-24} \, \text{g}\), the molecular mass is calculated by: \( \text{Molar mass of } X_2 = \frac{166.59 \times 10^{-24} \, \text{g}}{1 \times 6 \times 10^{23} } = 277.65 \, \text{g/mol} \).
5. Calculate the number of molecules in \(200 \, \text{g}\) of \(X_2\):
   • First, find the number of moles of \(X_2\) in \(200 \, \text{g}\): \( \text{Number of moles} = \frac{200 \, \text{g}}{277.65 \, \text{g/mol}} \approx 0.72 \, \text{mol} \).
   • Therefore, the number of molecules is: \( 0.72 \times 6 \times 10^{23} = 4.32 \times 10^{23} \approx 4 N_A \).

Hence, the number of molecules of \(X_2\) present in \(200 \, \text{g}\) is approximately \(4 \, N_A\), which matches the correct answer.