Animated Solution for Chemistry - States of Matter: A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of $400\text{ pm}$. If the density of the substance in the crystal is $8{\text{ g cm}}^{-3}$, then the number of atoms present in $256\text{ g}$ of the crystal is $N\times 10^{24}$. The value of N is :

Decoding the FCC Lattice

Imagine you are holding a perfectly ordered crystal in your hand. This crystal is made of tiny, repeating building blocks called unit cells. In this problem, we are told that the crystal has a Face-Centered Cubic (FCC) structure.

This means that if we look at a single unit cell, there is an atom at every corner of the cube, and an atom right in the middle of every face. The problem gives us three crucial pieces of information: the edge length of this unit cell ($a=400\text{ pm}$), the density of the crystal ($d=8{\text{ g/cm}}^3$), and the total mass of the crystal we are analyzing ($W=256\text{ g}$). Our mission is to find the total number of atoms in this $256\text{ g}$ sample.

The Smart Path

Most students, when they see density and edge length, immediately jump to the standard density formula:

They plug in the values to find the molar mass ($M$), and then use the mole concept to find the total number of atoms. While this method is perfectly correct, it is a long, winding road filled with large numbers and Avogadro's constant ($N_A$), which can easily lead to calculation errors.

But what if I told you there is a much faster, more elegant way? Let's think in terms of volumes.

Thinking in Volumes

Instead of worrying about moles, let's just figure out how much space our crystal takes up. We know the total mass and the density, so finding the total volume is a breeze:

Now, let's look at the microscopic level. How much space does a single unit cell occupy? The edge length is given as $400\text{ pm}$. Since our total volume is in cubic centimeters, we must convert this edge length to centimeters to keep our units consistent:

The volume of one cubic unit cell is simply the edge length cubed:

From Cells to Atoms

Now we have the total volume of the crystal and the volume of a single unit cell. To find out how many unit cells are packed into our crystal, we just divide the total volume by the volume of one cell:

We are almost at the finish line! We know we have $0.5\times 10^{24}$ unit cells. But the question asks for the total number of atoms. This is where the FCC structure comes into play. In an FCC lattice, the effective number of atoms per unit cell ($Z$) is 4.

Therefore, the total number of atoms is:

The problem states that the number of atoms is $N\times 10^{24}$. By comparing our result, it is crystal clear that $N=2$.

By thinking geometrically and using volumes, we completely bypassed Avogadro's number and arrived at the answer with simple, clean arithmetic!