Problem 7 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

7\. At \(100^{\circ} \mathrm{C}\), copper (Cu) has FCC unit cell structure with cell edge length of \(x \AA\). What is the approximate density of \(\mathrm{Cu}\) (in \(\left.\mathrm{g} \mathrm{cm}^{-3}\right)\) at this temperature? \(\begin{aligned}&\text { [Atomic Mass of } \mathrm{Cu}=63.55 \mathrm{u}] & \text }\end{aligned}\) (a) \(\frac{205}{x^{3}}\) (b) \(\frac{105}{x^{3}}\) (c) \(\frac{211}{x^{3}}\) (d) \(\frac{422}{x^{3}}\)

Step-by-Step Solution

Answer

The approximate density of Cu is \( \frac{211}{x^3} \) g/cm³, corresponding to option (c).

Step 1: Calculate Mass of Atoms in Unit Cell

Copper has a face-centered cubic (FCC) structure, where each unit cell contains 4 atoms. Given that the atomic mass of copper is 63.55 u (1 u = 1/NA g where NA = Avogadro's number \(6.022 \times 10^{23}\) mol\(^{-1}\)), we first calculate the mass of atoms in one unit cell. Thus, the mass of one atom of copper in grams is:\[ \text{Mass of one Cu atom} = \frac{63.55}{6.022 \times 10^{23}} \text{ g} \]So, the mass of 4 copper atoms in one unit cell is:\[ \text{Mass of 4 Cu atoms} = 4 \times \frac{63.55}{6.022 \times 10^{23}} \text{ g} \]

Step 2: Calculate Volume of Unit Cell

The volume of the unit cell can be calculated using the formula for volume of a cube since the FCC structure has cubic symmetry:\[ V = a^3 \]where \( a \) is the edge length of the cube. In this problem, it is given as \( x \mathring{A} \) where \( 1 \mathring{A} = 10^{-8} \text{ cm} \), so:\[ V = (x \times 10^{-8} \text{ cm})^3 = x^3 \times 10^{-24} \text{ cm}^3 \]

Step 3: Use Density Formula

The density \( \rho \) is obtained by dividing the total mass of atoms in the unit cell by the unit cell's volume:\[ \rho = \frac{\text{Mass of 4 Cu atoms}}{\text{Volume of Unit Cell}} = \frac{4 \times \frac{63.55}{6.022 \times 10^{23}}}{x^3 \times 10^{-24}} \text{ g/cm}^3 \]

Step 4: Simplify the Expression

Simplify the expression for density computed in the previous step:Substituting the value of Avogadro's number \( N_A = 6.022 \times 10^{23} \) gives:\[ \rho = \frac{4 \times 63.55}{6.022 \times 10^{23}} \times \frac{1}{x^3 \times 10^{-24}} \]Simplifying further:\[ \rho = \frac{4 \times 63.55 \times 10^{24}}{6.022 \times x^3} \]\[ \rho = \frac{254.20 \times 10^{24}}{6.022 \times x^3} \]Hence the approximate density of Cu at this temperature can be approximated as:\[ \rho \approx \frac{211}{x^3} \text{ g/cm}^3 \]Therefore, the correct choice is (c).

Key Concepts

Face-Centered Cubic Structure

In solid-state chemistry, the structure of a crystal significantly influences its physical properties. One common crystal structure is the Face-Centered Cubic (FCC) structure.
In an FCC crystal, the atoms are arranged in a cubic lattice. Each cube has atoms positioned at each of its corners and an additional atom centered in each face of the cube. This dense packing allows each atom to be surrounded by 12 other atoms, maximizing close-contact bonding and characteristic stability.

• Each corner atom is shared among eight adjacent unit cells, contributing only 1/8 of its mass to a given cell.
• Each face-centered atom is shared between two unit cells, contributing half of its mass to the unit cell.

By this arrangement, a single FCC unit cell contains a total of 4 whole atoms. Understanding this structure helps us calculate properties like density, because knowing the number of atoms per unit cell is crucial for determining the mass within that cell.

Atomic Mass

The atomic mass of an element is a fundamental property indicating its relative weight on a scale preferred for atoms. Measured in atomic mass units (amu or u), it represents the average mass of an atom of an element, accounting for the various isotopes and their abundances.
For copper (Cu), the atomic mass is 63.55 u, translating to grams when needed for calculations, such as density computations. We use Avogadro's constant to convert from amu to grams, reflecting the mass of many atoms rather than a solitary one.
Mathematically, the atomic mass in grams for a mole of atoms (often termed molar mass) guides the translation from microscopic calculations to macroscopic properties:

• Convert atomic mass units to grams using the relationship: 1 u = 1 g/mol.
• Use copper's atomic mass in density calculations by multiplying atomic mass by the number of atoms per unit cell.

This approach ties the world of single atoms, which are abstract and imaginary, to tangible quantities that we can see and measure in experimental work.

Avogadro's Number

Avogadro's number (\(N_A\)) is a pivotal constant in chemistry and physics, representing the number of atoms, molecules, or particles found in one mole of substance. This constant is valued at approximately \(6.022 \times 10^{23}\), providing a bridge between atomic-scale computations and real-world quantities.
Avogadro's number allows us to effectively use atomic masses, measured in amu, into a usable weight in grams. When calculating quantities like density, Avogadro's number converts mass from an atomic to a molar scale and vice-versa.
Key uses include:

• Translating atomic mass units (amu) to grams for density calculations.
• Allowing straightforward conversion from moles to particles, facilitating the comprehension of amounts involved in chemical reactions.

Its regular use in calculations helps unify microscopic and macroscopic perspectives, reinforcing our understanding of material properties and reactions in chemistry and physics.