Question

A diatomic molecule $X_2$ has a body-centred cubic ($\mathrm{bcc}$) structure with a cell edge of $300pm$. The density of the molecule is $6.17g/c{m}^{-1}$. The number of molecules present in $200g$ of $X_2$ is: (Avogadro constant $\left(N_A\right)=6.023\times {10}^{23}$)

• A. $8N_A$
• B. $2N_A$
• C. $40N_A$
• D. $4N_A$

Answer 1

The correct option is D

$4N_A$

The above question can be solved as below:

Step 1: It is given that density $\rho =6.17g/c{m}^3$,

$a=300pm$,

$N_A=6.023\times {10}^{23}$,

Step 2: The formula for $$\begin{gathered} \rho =\frac{2\times {\displaystyle \frac{GMM}{N_A}}}{a^3} \\ =\frac{2\times GMM}{N_A\times a^3} \end{gathered}$$

Step 3: Reforming the equation to find the number of molecules: $GMM=\frac{\rho \times a^3\times N_A}{2}$

Step 4: Putting the values in the equation and calculating: $$\begin{gathered} GMM=\frac{6.17\times {\left(300\times {10}^{-10}\right)}^3\times 6.023\times {10}^{23}}{2} \\ =\frac{6.17\times 6\times 9\times 3\times {10}^{-1}}{2} \\ GMM=81\times 6.17\times {10}^{-1} \\ =49.97g/mol \end{gathered}$$

Step 5: The number of molecules present in $1mole$ of the diatomic molecule is $49.97g/mol$.

Step 6: The number of molecules in $200g$ of diatomic molecule $$\begin{gathered} =\frac{200}{49.47}\times N_{A}g/mol \\ \cong \mathbf{4}\mathbf{}{\mathit{N}}_{\mathbf{A}}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

The number of molecules present in $200g$ of $X_2$ is $4{\mathrm{N}}_{\mathrm{A}}\mathrm{g}/\mathrm{mol}$