Question

Gold crystallizes with FCC lattice for which the side length of the unit cell is \(5.0 \AA\). If the density of gold is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\) the value of Avogadro number is \((\mathrm{Au}=198)\) (a) \(6.022 \times 10^{23}\) (b) \(6.034 \times 10^{23}\) (c) \(5.966 \times 10^{23}\) (d) \(6.022 \times 10^{22}\)

Short answer

\(6.022 \times 10^{23}\)

Step-by-step solution

Understanding the Problem

The task is to calculate Avogadro's number given the density of gold, the edge length of the face-centered cubic (FCC) lattice, and the atomic mass of gold. The formula for calculating the number of atoms in an FCC unit cell is required, along with the density formula to relate mass, volume, and the number of atoms.

Calculate the Number of Atoms in the FCC Unit Cell

An FCC unit cell has 4 atoms per unit cell. This calculation does not need to consider the atomic radius as it is not provided or necessary for this formula.

Convert Edge Length from Angstroms to Centimeters

The edge length is given in Angstroms (\r{A}) where 1 \r{A} = 1 \(\times 10^{-8}\) cm. So, \(5.0 \r{A}\) = \(5.0 \times 10^{-8}\) cm.

Calculate the Volume of the Unit Cell

The volume (V) of the unit cell is the cube of the edge length: \(V = a^3\), where \(a\) is the edge length. Calculate \(V\) using \(a = 5.0 \times 10^{-8}\) cm.

Calculate the Mass of One Unit Cell

The mass of one unit cell can be calculated using the density formula: \(\rho = \frac{m}{V}\). Rearrange the formula to solve for mass (m): \(m = \rho \times V\).

Determine the Mass of One Atom of Gold

The mass of one atom of gold (M_atom) is the mass of one unit cell divided by the number of atoms in the cell (4).

Calculate Avogadro's Number

Avogadro's number (N_A) is the ratio of the molar mass (Molar_mass) of gold to the mass of one atom of gold (M_atom): \(N_A = \frac{Molar_mass}{M_atom}\). The molar mass of gold (Au) is 198 g/mol.

Verify the Answer

Use the calculated value of Avogadro's number to compare with the given options and determine the correct one.

Key concepts

Face-Centered Cubic Lattice

A face-centered cubic (FCC) lattice is a type of crystal structure where atoms are located at each corner and the centers of all the faces of the cube. This arrangement is significant in materials science because it can be found in many metals, including gold. In an FCC lattice, each unit cell contains 4 atoms in total: one-eighth of an atom from each of the eight corners (1 atom total) and one-half of an atom from each of the six face-centers (3 atoms total). This unique configuration leads to high packing efficiency, which means the atoms are closely packed together, occupying approximately 74% of the space in the lattice.

Understanding the arrangement of atoms in this lattice is crucial as it affects the physical properties of the material, such as its density and strength. Additionally, it helps to comprehend how the atoms interact in the lattice and how altering the structure impacts the material's properties.

Unit Cell Volume

The volume of a unit cell is a fundamental concept in the study of crystalline structures. It is defined as the cubic space occupied by a unit cell and is calculated from the edge length of the cell. Specifically, for a face-centered cubic lattice, the volume (V) is determined by cubing the edge length (a), resulting in the formula V = a^3.

Unit cell volume is directly related to other physical properties such as density and atomic packing. In our example with a gold crystal lattice, converting the given edge length from Angstroms to centimeters is an essential step before volume calculation. The detailed understanding of unit cell volume is necessary for determining the density of the crystal and for handling calculations that involve Avogadro's number, which represents the number of entities in a mole of a substance.

Atomic Mass

Atomic mass, often termed atomic weight, refers to the mass of a single atom and is typically expressed in atomic mass units (amu) or grams per mole as molar mass. For elements like gold (Au) that appear in the periodic table, the atomic mass is usually the average mass of all isotopes, weighted by their relative abundance. However, in practical scenarios, like calculating the Avogadro number, we use the molar mass, which is the mass of one mole of atoms.

In contexts of crystalline structures, knowing the atomic mass allows us to relate a single atom's mass to the entire sample's mass. For instance, in the case of gold, with a molar mass of 198 g/mol, it's possible to determine the mass of a single gold atom by dividing this value by Avogadro's number. This provides a pivotal piece of information when we aim to calculate the density of a substance and ultimately contributes to comprehension of material properties.

Crystal Density

Crystal density is a measure of how tightly the atoms are packed within a crystal lattice. It is calculated as the mass of the crystal divided by its volume. In the context of a face-centered cubic lattice, where the unit cell volume can be determined based on the cube of the edge length, the crystal density tells us how much mass is present in a given volume of the crystal.

When calculated as ρ = m/V, where m is mass and V is volume, the crystal density provides insight into the nature of the material. In case of gold with a high density, this reflects a tightly packed atomic arrangement. Crystal density is not only vital in materials science but also in understanding various physical and chemical properties of substances, such as buoyancy and purity. These properties are essential in diverse applications, including jewel crafting where gold’s high density contributes to its value and heft.