Animated Solution for Chemistry - States of Matter: Ga (atomic mass 70 u) crystallises in a hexagonal close packed structure. The total number of voids in 0.581 g of Ga is ...... $\times 10^{21}$. (Round off to the nearest integer).

The journey into the microscopic world of solid-state chemistry is nothing short of fascinating. When we look at a piece of metal like Gallium, it appears as a continuous, solid mass. However, at the atomic level, it is a highly organized, bustling metropolis of atoms packed together in specific geometric patterns. In this problem, we are dealing with the Hexagonal Close Packed (HCP) structure.

Our mission is to find the total number of empty spaces—known as voids—hidden within a tiny $0.581\text{ g}$ sample of Gallium. Let's break down the physics and mathematics behind this beautiful problem step by step.

Imagine packing oranges in a crate. You want to fit as many as possible. Nature does the exact same thing with atoms. In an HCP structure, atoms are arranged in alternating layers (A-B-A-B), creating a highly dense packing arrangement.

For any unit cell, the most critical parameter is $Z$, the effective number of atoms. For an HCP unit cell, the math works out perfectly to give us:

But here is the catch: no matter how tightly you pack spheres, there will always be empty gaps between them. These gaps are called voids. There are two primary types of voids in close-packed structures: 1. Tetrahedral Voids (TV): Formed between four atoms. 2. Octahedral Voids (OV): Formed between six atoms.

There is a beautiful, universal geometric rule in solid-state chemistry that links the number of atoms to the number of voids. For any close-packed lattice with $Z$ effective atoms: - The number of Octahedral Voids is exactly equal to $Z$. - The number of Tetrahedral Voids is exactly double, $2Z$.

Applying this to our HCP unit cell:

If we add these together, the total number of voids in a single HCP unit cell is:

This gives us a profound insight: for every $6$ atoms in the lattice, there are $18$ voids. This means the ratio of voids to atoms is exactly $3:1$. Every single atom effectively brings $3$ voids into existence!

The Mole Concept

Now that we understand the geometry, let's bring in the chemistry. We have a $0.581\text{ g}$ sample of Gallium, and its atomic mass is $70\text{ u}$.

First, we need to find out how many moles of Gallium we are dealing with. The formula is straightforward:

Substituting our values:

To find the total number of atoms, we multiply the moles by Avogadro's number ($N_A$):

We know that the total number of voids is simply $3$ times the total number of atoms. Let's set up our master equation:

Let's do the arithmetic carefully.

So, our expression becomes:

To match the format requested in the question ($\times 10^{21}$), we shift the decimal point two places to the right:

The question asks us to round off our answer to the nearest integer. Looking at $14.99$, it is incredibly close to $15$.

Therefore, rounding off gives us our final, elegant answer:

This problem is a perfect symphony of 3D geometry and stoichiometry. By understanding the fundamental ratio of atoms to voids, we bypassed complex unit cell calculations and arrived at the answer with pure logical deduction. Always look for these hidden ratios—they are the secret weapons of elite problem solvers!