Animated Solution for Chemistry - Solid State: A diatomic molecule $X_2$ has a body-centred cubic (bcc) structure with a cell edge of $300\text{ pm}$. The density of the molecule is $6.17{\text{ g cm}}^{-3}$. The number of molecules present in $200\text{ g}$ of $X_2$ is (Avogadro constant $(N_A)=6\times 10^{23}{\text{ mol}}^{-1}$)

Imagine you are zooming into the microscopic world of a crystal lattice. We are given a diatomic molecule, $X_2$, that crystallizes in a body-centered cubic (bcc) structure.

In a bcc unit cell, molecules are positioned at all eight corners of the cube, and there is exactly one molecule sitting right in the center of the body. Because corner molecules are shared among eight adjacent unit cells, their total contribution is $8\times \frac{1}{8}=1$. Adding the central molecule, the effective number of molecules per unit cell, denoted as $Z$, is exactly $2$.

We are also provided with the edge length of the unit cell, $a=300\text{ pm}$, and the density of the crystal, $d=6.17{\text{ g cm}}^{-3}$. Our ultimate goal is to find the total number of molecules present in a $200\text{ g}$ sample of this substance.

To bridge the gap between the microscopic unit cell and the macroscopic mass, we need to find the molar mass ($M$) of the substance. The master equation that connects these properties is the density formula for a crystal lattice:

Before we substitute our values, we must ensure our units are consistent. The density is given in ${\text{g cm}}^{-3}$, so our edge length must be converted from picometers to centimeters:

Now, let's substitute the known values into our density equation:

Let's carefully expand the denominator to avoid any silly mistakes with the powers of ten. Cubing the edge length gives:

Multiplying this volume by Avogadro's number ($N_A$):

Now our equation looks much cleaner and less intimidating:

Rearranging to solve for the molar mass $M$:

We have successfully found that the molar mass of the diatomic molecule $X_2$ is $50\text{ g/mol}$. But the question asks for the total number of molecules in a $200\text{ g}$ sample.

First, we calculate the number of moles ($n$) in the sample by dividing the given mass ($w$) by the molar mass ($M$):

Finally, to find the total number of molecules ($N$), we simply multiply the number of moles by Avogadro's number ($N_A$):

This perfectly matches option (d). Always remember to double-check the lattice type in such problems, as a simple change from BCC to FCC would drastically alter the $Z$ value and the final result!