Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

At room temperature, sodium crystallizes in a BCC lattice with the cell edge (a) \(4.24 \AA\). Find the density of sodium. (Atomic wt of \(\mathrm{Na}=23\) ) (a) \(2.01 \mathrm{~g} / \mathrm{cm}^{3}\) (b) \(1.002 \mathrm{~g} / \mathrm{cm}^{3}\) (c) \(3.003 \mathrm{~g} / \mathrm{cm}^{3}\) (d) \(2.004 \mathrm{~g} / \mathrm{cm}^{3}\)

Step-by-Step Solution

Answer

The density of sodium is approximately \(1.002 \ \text{g/cm}^3\), which matches option (b).

Step 1: Identify Crystal Structure

Sodium crystallizes in a body-centered cubic (BCC) lattice. In a BCC structure, there are 2 atoms per unit cell.

Step 2: Convert Cell Edge to Centimeters

The cell edge of sodium is given as \(4.24 \ \text{Å}\). Convert this into centimeters: \(4.24 \ \text{Å} = 4.24 \times 10^{-8} \ \text{cm}\).

Step 3: Calculate Volume of Unit Cell

Compute the volume of the cubic unit cell using the formula for the volume of a cube \(a^3\), where \(a\) is the edge length.\[\text{Volume} = (4.24 \times 10^{-8} \ \text{cm})^3 = 7.61 \times 10^{-23} \ \text{cm}^3\].

Step 4: Determine Mass of Unit Cell

The number of atoms per unit cell in BCC is 2. Using Avogadro's number \(6.022 \times 10^{23} \text{ atoms/mole}\), find the mass of the unit cell:\[\text{Mass} = \left(\frac{2 \text{ atoms} \times 23 \text{ g/mole}}{6.022 \times 10^{23} \text{ atoms/mole}}\right) = 7.64 \times 10^{-23} \ \text{g}\].

Step 5: Calculate Density

Use the density formula \( \text{Density} = \frac{\text{mass}}{\text{volume}} \). Substitute the values obtained:\[\text{Density} = \frac{7.64 \times 10^{-23} \ \text{g}}{7.61 \times 10^{-23} \ \text{cm}^3} \approx 1.0045 \ \text{g/cm}^3\].

Step 6: Compare Calculated Density to Options

Compare the calculated density \(1.0045 \ \text{g/cm}^3\) with the provided options. It is closest to option (b) \(1.002 \ \text{g/cm}^3\).

Key Concepts

Body-Centered Cubic (BCC) Lattice

The Body-Centered Cubic (BCC) lattice is one of the fundamental ways that atoms can be arranged in a crystalline structure. In this setup, each unit cell, which is the smallest repeatable unit in a crystal, includes an atom at each corner and a single atom in the very center.
This structural arrangement is significant because:

• It has 2 atoms per unit cell: 8 corner atoms each contribute 1/8 of an atom plus the 1 atom in the center.
• The BCC lattice is denser than simple cubic lattices but less dense than face-centered cubic (FCC) lattices.

Understanding BCC lattices is crucial as it influences how the material behaves, for instance, its density and atomic interactions. Sodium, among other metals like iron at room temperature, crystallizes in this manner, providing it with unique physical properties.

Unit Cell Calculations

Unit cell calculations involve determining values like volume and mass that are fundamental to finding the density of a substance. The unit cell is a cube if the lattice is cubic, which is often the case for metals. Calculating the volume is straightforward:

• You simply use the formula for the volume of a cube: \(a^3\), where \(a\) is the cell edge length.
• The cell edge length needs to be in consistent units, typically centimeters, when calculating density.

For sodium, with an edge length of \(4.24 \ \text{Å} \), converting to centimeters is necessary for further calculations. This conversion ensures that all measurements align, simplifying the process of calculating mass and density.

Sodium Crystal Structure

Sodium is a fascinating element because it adopts a BCC crystal structure at room temperature. This structure impacts its properties, such as density, melting point, and electrical conductivity.

• The BCC structure of sodium allows it to have a lower density compared to other configurations, despite being a solid metal.
• The arrangement of atoms in sodium means that it is more reactive than metals with more compact atomic arrangements.

As sodium crystallizes into a BCC lattice, we see how its atomic structure directly translates into its physical properties. This understanding helps chemists and material scientists predict behavior in applications ranging from simple table salt to complex reactors.

Atomic Weight and Avogadro's Number

Atomic weight and Avogadro's number are two vital concepts in chemistry that assist in the calculation of quantities like mass in larger scales.

• Atomic weight, expressed usually in grams per mole, tells us the mass of 1 mole of an element – for sodium, it's 23 g/mol.
• Avogadro's number, \(6.022 \times 10^{23}\) atoms/mole, reveals the number of atoms in one mole of any substance, establishing a link between atomic-scale and macroscopic quantities.

When calculating properties like the density of a crystal, these concepts help us transition from a small number of atoms in a unit cell to the actual weight of the element contained in that unit cell. By knowing the atomic weight and using Avogadro's number, students can comprehensively understand the quantity of substance engaged in reactions or in solid states.