Problem 17 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A metal having atomic mass \(60.22\) \(\mathrm{g} /\) mole crystallizes in \(\mathrm{ABCABC} \ldots\) type packing. The density of each metal atom if the edge length of unit cell is \(10 \AA\), is \(\left(N_{\mathrm{A}}=6.022 \times 10^{23}\right)\) (a) \(0.4 \mathrm{~g} / \mathrm{cm}^{3}\) (b) \(40 \mathrm{~g} / \mathrm{cm}^{3}\) (c) \(0.54 \mathrm{~g} / \mathrm{cm}^{3}\) (d) \(54 \mathrm{~g} / \mathrm{cm}^{3}\)

Step-by-Step Solution

Answer

54 \(\mathrm{g}/\mathrm{cm}^3\)

Step 1: Determine the Packing Type and the Number of Atoms per Unit Cell

Since the metal crystallizes in ABCABC type packing, which is also known as cubic close-packed (ccp) or face-centered cubic (fcc) packing, each unit cell contains 4 atoms of the metal.

Step 2: Calculate the Volume of the Unit Cell

The edge length of the unit cell is given as 10 Å (angstroms). Since volume is equal to the cube of the edge length, the volume of the unit cell is \(10 \text{ Å}\)^3 = (10 \times 10^{-10} \text{m})^3 = 10^{-27} \text{m}^3. Since 1 \text{cm}^3 is equal to 10^{-6} \text{m}^3, the volume in \text{cm}^3 is 10^{-21} \text{cm}^3.

Step 3: Calculate the Mass of Atoms in One Unit Cell

Since there are 4 atoms in each unit cell for fcc packing and the atomic mass is given as 60.22 g/mol, the mass of 4 atoms is \(\frac{4 \times 60.22 \text{ g}}{N_{\text{A}}}\), where \(N_{\text{A}} = 6.022 \times 10^{23} \) is Avogadro's number.

Step 4: Determine the Mass of One Atom

The mass of one atom is \(\frac{60.22 \text{ g/mol}}{6.022 \times 10^{23} \) atoms/mol}, which gives \(10^{-22}\text{ g}\) per atom approximately.

Step 5: Calculate the Density of the Metal Atom

Density \(\rho\) is mass per unit volume. For the unit cell the density would be \(\frac{4 \times \text{mass of one atom}}{\text{volume of the unit cell}}\). Using the values from previous steps, plug in the mass of one atom and the volume of the unit cell to get the density.

Key Concepts

Cubic Close-Packed Structure

Understanding the crystal structure of materials is critical to predicting their properties. In the cubic close-packed (ccp) structure, which is also referred to as face-centered cubic (fcc), atoms are arranged in a way that each atom is surrounded by twelve others, forming a pattern with high packing efficiency. Imagine stacking layers of spheres: the bottom layer is the 'A' layer, where each sphere touches four neighbors. The next layer, 'B', fits into the gaps of 'A'. Then 'C' layer spheres go into the alternate gaps of the 'B' layer that aren't directly above 'A'. This pattern keeps repeating, which is why we describe it as 'ABCABC...'.

For many metals, the ccp structure is favored because it allows the maximum fill of space with the minimum of voids, leading to high density. When it comes to calculating density, this repeating sequence has particular implications: it dictates not only how atoms align, but also how many atoms can fit within a unit cell – for ccp, the answer is 4 atoms per unit cell.

Unit Cell Volume

When we delve into the microscopic world to calculate properties like density, we often refer to the concept of a unit cell – the smallest repeating unit that shows the full symmetry of the crystal structure. For a cubic structure, calculating the volume of the unit cell is straightforward; it's simply the cube of the edge length of the cell. The calculation for the volume of a unit cell in the exercise at hand involves cubing the given edge length (10 Å), and then converting the cubic angstroms into cubic centimeters, as the latter is a more standard unit for expressing density.

To facilitate understanding, think of the cubic unit cell as a tiny box whose edges are as long as the given measurement. The internal space of this box (its volume) represents where atoms can fit, and thus plays a central role when we calculate the density of the crystal. In this case, the unit cell's volume helps us determine how much mass is contained within a defined space – a fundamental step in the process of finding the material's density.

Avogadro's Number

A concept vital to chemistry and physics, and indeed to understanding the nature of matter on a molecular level, is Avogadro's number. Named after the scientist Amedeo Avogadro, this number defines the amount of constituent particles (usually atoms or molecules) in one mole of a substance; its value is approximately 6.022 x 10^23. This not-so-arbitrary number allows us to link the macroscopic world of grams and centimeters with the atomic-scale world.

For the exercise in question, Avogadro's number is key to connecting the atomic mass of the metal (given in grams per mole) to the mass of individual atoms needed for calculating density. Crucial for Step 3 and Step 4 of the solution, this number allows for the conversion from the molar mass of the substance to the mass of one atom and then, combining with the volume of the unit cell, lets us determine the density of the substance in more familiar terms. It serves as a bridge between the atomic properties and the tangible measurements we make in the lab or industry.