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

Question

At room temperature, sodium crystallizes in a \(\mathrm{BCC}\) 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 \(1.002 \mathrm{~g/cm}^3\).

Step 1: Identify Structure and Properties

In this problem, sodium crystallizes in a body-centered cubic (BCC) structure. In a BCC unit cell, there are two atoms per unit cell.

Step 2: Determine Volume of Unit Cell

The cell edge length is provided as \( a = 4.24\, \AA \). Convert this to cm: \( a = 4.24 \times 10^{-8} \mathrm{~cm} \). The volume of the unit cell \( V \) can be calculated as \( V = a^3 \). Therefore, \( V = (4.24 \times 10^{-8} \mathrm{~cm})^3 = 7.61 \times 10^{-23} \mathrm{~cm}^3 \).

Step 3: Calculate Mass of Unit Cell

The atomic weight of sodium is \( 23 \mathrm{~g/mol} \). Use Avogadro's number \( 6.022 \times 10^{23} \mathrm{~atoms/mol} \) to find the mass of two atoms: \[ \text{Mass of unit cell} = 2 \times \left( \frac{23 \mathrm{~g/mol}}{6.022 \times 10^{23} \mathrm{~atoms/mol}} \right) = 7.64 \times 10^{-23} \mathrm{~g} \].

Step 4: Calculate Density

Density \( \rho \) is the mass per unit volume: \[ \rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{7.64 \times 10^{-23} \mathrm{~g}}{7.61 \times 10^{-23} \mathrm{~cm}^{3}} = 1.004 \mathrm{~g/cm}^3 \].

Step 5: Choose the Correct Answer

Comparing our calculated density (\(1.004 \mathrm{~g/cm}^3\)) to the provided options, the closest match is option (b) \(1.002 \mathrm{~g/cm}^3\).

Key Concepts

BCC lattice

Sodium in its solid form at room temperature adopts a body-centered cubic (BCC) structure. This particular crystal structure is one of the basic types in crystallography and is known for its simple geometry. In a BCC lattice, each unit cell contains an atom at each corner of the cube and one atom in the center, totaling two atoms per unit cell.
The atoms at the corners are shared between adjoining cells. Therefore, each corner atom contributes one-eighth to the unit cell. The atom in the center is fully contained within the cell, straightforwardly accounting for the integral contribution of one atom. The BCC structure is an efficient way for sodium atoms to pack together due to their size and electronic configuration.

unit cell volume

The concept of unit cell volume is crucial for determining the properties of crystalline solids like density. For a cube, which is the shape of the simplest unit cell, the volume can be calculated as the cube of the edge length. Given the edge length of a sodium BCC lattice as 4.24 Å (angstroms), we need to convert it to centimeters for practical calculations.
Converting to centimeters involves multiplying by a factor of \(10^{-8}\), making the edge length 4.24 × 10^{-8} cm. The volume of the unit cell is then calculated by cubing this value, \[V = (4.24 \times 10^{-8} \, \text{cm})^3 = 7.61 \times 10^{-23} \, \text{cm}^3\].
This tiny volume reflects the scale at which atomic arrangements impact macroscopic material properties.

atomic weight

Atomic weight, often expressed in grams per mole (g/mol), represents the average mass of an element's atoms compared to 1/12th the mass of a carbon-12 atom. For sodium, the atomic weight is given as 23 g/mol. This value is used to translate the conceptual mass of a mole of sodium atoms into a practical weight.
When calculating properties like density, atomic weight helps to determine the mass of atoms contained in a unit cell. It provides a crucial conversion between atomic and macroscopic levels, connecting the tiny mass of individual atoms to the larger-scale quantities used in experiments and applications.
In our problem, understanding the atomic weight allows us to relate Avogadro's number of atoms to the mass of a single unit cell.

Avogadro's number

Avogadro's number is a fundamental constant in chemistry representing the number of atoms, ions, or molecules in one mole of substance. Its value, approximately \(6.022 \times 10^{23}\), allows chemists to count entities at the atomic scale by weighing measurable amounts of material.
In crystallography and density calculations, Avogadro's number links the atomic scale to the macroscopic scale. For example, in finding the mass of a unit cell, we multiply the atomic weight by the number of atoms in the unit cell divided by Avogadro's number. This gives the mass that can be expected if we could isolate and weigh such a tiny amount.
In essence, Avogadro's number is a bridge that connects the atoms' world to our everyday world, enabling practical applications of atomic theory.