Problem 111 - A Complete Resource Book in Chemistry for JEE Main

Question

Gold (Au) crystallizes in cubic close packed (FCC). The atomic radius of gold is \(144 \mathrm{pm}\) and the atomic mass of \(\mathrm{Au}=197.0 \mathrm{amu}\). The density of \(\mathrm{Au}\) is (a) \(19.4 \mathrm{~g} \mathrm{~cm}^{-3}\) (b) \(194 \mathrm{~g} \mathrm{~cm}^{-3}\) (c) \(39.4 \mathrm{~g} \mathrm{~cm}^{-3}\) (d) \(0.194 \mathrm{~g} \mathrm{~cm}^{-3}\)

Step-by-Step Solution

Answer

The density of gold is closest to (a) 19.4 g/cm³.

Step 1: Determine the Edge Length of the Unit Cell

A face-centered cubic (FCC) unit cell has four atoms. In an FCC structure, the edge length \( a \) is related to the atomic radius \( r \) by the equation \( a = 2\sqrt{2}\times r \). Given \( r = 144 \mathrm{pm} \), convert this to centimeters: \( r = 144 \times 10^{-10} \mathrm{cm} \). Calculate \( a \): \( a = 2\sqrt{2}\times 144 \times 10^{-10} \mathrm{cm} \approx 4.08 \times 10^{-8} \mathrm{cm} \).

Step 2: Calculate the Volume of the Unit Cell

The volume of a cubic unit cell is given by \( V = a^3 \). From Step 1, \( a \approx 4.08 \times 10^{-8} \mathrm{cm} \). Thus, \( V = (4.08 \times 10^{-8})^3 \mathrm{cm}^3 \approx 6.79 \times 10^{-23} \mathrm{cm}^3 \).

Step 3: Find the Mass of the Unit Cell

The mass of the unit cell can be determined by knowing that there are 4 atoms in a FCC unit cell. Each atom has a mass given by the molar mass divided by Avogadro's number: \( m = \frac{197.0 \mathrm{g/mol}}{6.022 \times 10^{23}/\text{mol}} \approx 3.27 \times 10^{-22} \mathrm{g} \). Total mass of the unit cell is \( 4 \times 3.27 \times 10^{-22} \mathrm{g} \approx 1.31 \times 10^{-21} \mathrm{g} \).

Step 4: Calculate the Density of Gold

The density \( \rho \) is the mass of the unit cell divided by its volume. From Step 3 and Step 2, we have \( \rho = \frac{1.31 \times 10^{-21} \mathrm{g}}{6.79 \times 10^{-23} \mathrm{cm}^3} \approx 19.3 \mathrm{g/cm}^3 \).

Step 5: Choose the Closest Answer from the Options

The density calculated is approximately \( 19.3 \mathrm{g/cm}^3 \). Comparing with the options, the closest answer is (a) \( 19.4 \mathrm{g/cm}^3 \).

Key Concepts

face-centered cubic (FCC)

The face-centered cubic (FCC) crystal structure is a highly efficient packing arrangement where atoms are located at each of the cube's corners and the centers of each face. This structure is also referred to as cubic close-packed (CCP). Each unit cell in an FCC lattice contains a total of four atoms. The FCC arrangement is significant because it offers one of the highest packing efficiencies at 74%, meaning the atoms are packed very closely together. In metals like gold, which follow this structure, this efficient packing contributes to their density and physical properties. The special feature of FCC is that despite the close packing, there are significant voids between atoms, allowing for malleability and ductility in metals with this structure.

atomic radius

The atomic radius is a measure of the size of an atom, typically the distance from the nucleus to the outermost shell of electrons. It is directly related to how atoms pack in a crystal structure.For face-centered cubic (FCC) crystals like gold, the atomic radius helps define the length of the cube edges. The relationship is specifically given by the formula: \[ a = 2\sqrt{2}\times r \]where \( a \) is the edge length of the cube, and \( r \) is the atomic radius. Gold has an atomic radius of 144 pm (picometers). Calculating the edge length from the radius helps in determining the volume of the unit cell, which is crucial for computing density.

Avogadro's number

Avogadro's number is a fundamental quantity in chemistry and physics, defined as the number of constituent particles, usually atoms or molecules, contained in one mole of a substance. It is approximately equal to \(6.022 \times 10^{23}\) particles/mol.This number serves as a bridge between the atomic scale and macroscopic quantities, allowing chemists and physicists to relate atomic mass to measurable quantities like grams. In calculating the mass of a gold atom as part of its FCC unit cell, Avogadro's number is used to convert between the molar mass of gold and the mass of a single atom:\[ m = \frac{197.0 \mathrm{g/mol}}{6.022 \times 10^{23}/\text{mol}} \approx 3.27 \times 10^{-22} \mathrm{g} \]Through this conversion, it becomes possible to find the mass of any number of atoms in a specified volume of a material.

molar mass

Molar mass is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). In practice, it serves as an essential factor in converting between mass and the number of atoms or molecules. For gold, the molar mass is 197.0 g/mol. In calculations involving the face-centered cubic (FCC) structure, the molar mass helps in finding individual atomic mass by utilizing Avogadro's number. To calculate the density of a substance like gold, knowing the molar mass allows one to compute the mass of individual atoms and subsequently determine properties like mass of a unit cell. For example, the mass of gold atoms within an FCC unit cell is determined by multiplying the number of atoms in the cell by the individual atomic mass, enabling calculations of density and closely related physical properties.