Problem 26 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of \(400 \mathrm{pm}\). If the density of the substance in the crystal is \(8 \mathrm{~g} \mathrm{~cm}^{-3}\), then the number of atoms present in \(256 \mathrm{~g}\) of the crystal is \(N \times 10^{24}\). The value of \(N\) is ____ .

Step-by-Step Solution

Answer

The value of \(N\) is 4.

Step 1: Understanding the Structure

A face-centered cubic (FCC) structure means that there are atoms at each corner and at the center of each face of the cube. This type of structure contains 4 atoms per unit cell.

Step 2: Calculate Volume of Unit Cell

The edge length of the cube is given as 400 pm, which is equivalent to \(400 \times 10^{-10} \text{cm}\). The volume of the unit cell is given by \( (400 \times 10^{-10} \text{cm})^3 \).

Step 3: Calculate Mass of Unit Cell

Using the formula for density, \( \text{density} = \frac{\text{mass}}{\text{volume}} \), rearrange to find volume as \( \text{Volume} = \frac{\text{mass}}{\text{density}} \) and plug in the given density to find the mass of the unit cell: \( m_{cell} = 8 \text{g/cm}^3 \times \left( 400 \times 10^{-10} \right)^3 \text{cm}^3 \).

Step 4: Calculate Number of Atoms in 256 grams

Use the molar mass of the substance and Avogadro's number to convert mass to atoms. Start by dividing the total mass (256 g) by the mass of one unit cell, then multiply by the number of atoms per unit cell (4) to get the total number of atoms. Finally, express the number of atoms in the form \(N \times 10^{24}\) and solve for \(N\).

Step 5: Calculate Molar Mass of the Substance

Recognize that the mass of one unit cell and Avogadro's number will help us derive the molar mass. Given the formula \( \text{Molar mass} = \frac{\text{mass of unit cell} \times N_{A}}{n}\), solve for the molar mass.

Step 6: Solve for N

Calculate \(N\) using the expression obtained in the step above, such that the result satisfies \( 256 \text{g} = N \times 10^{24} \text{atoms} \).

Key Concepts

Density Calculation

Density is a critical property that describes how much mass is concentrated within a given volume. In the context of a face-centered cubic (FCC) structure, density can help us understand the molecular arrangement in a crystal solid. The density formula is commonly expressed as:

• \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)

For density calculation, we rearrange this to determine the mass of the unit cell:

• \( \text{Mass} = \text{Density} \times \text{Volume} \)

This concept is crucial for calculating how much substance is packed in a crystal lattice. It also allows us to deduce other important parameters such as molar mass or the number of atoms, given specific density information.
Understanding density in this context helps us connect mass and volume to the atomic structure of the material.

Unit Cell Volume

The unit cell is the smallest repeating unit in a crystal lattice that defines the entire structure by stacking in three-dimensional space. Calculating the unit cell volume involves using the edge length of the cubic cell, which in the given exercise is 400 pm. It's essential to convert these measurements into consistent units to ensure accuracy. Thus, we convert picometers to centimeters:

• \( 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} \)

The formula for the volume of a cube is:

• \( \text{Volume} = \text{Edge length}^3 \)

Substituting the edge length in its converted form, we determine the unit cell volume is:

• \( \left( 400 \times 10^{-10} \right)^3 \text{ cm}^3 \)

This calculation is fundamental for connecting physical dimensions with the properties of materials, particularly when evaluating density and molar mass.

Avogadro's Number

Avogadro's Number is a fundamental constant used in chemistry to relate macroscopic quantities of substances to the atomic scale. It is defined as the number of constituent particles, usually atoms or molecules, contained in one mole of a substance, approximately \(6.022 \times 10^{23}\). When connecting macroscopic mass to atomic numbers, Avogadro's Number is invaluable.
In our exercise, Avogadro's Number assists in finding the molar mass and the number of atoms within a specific mass of crystalline solid. Knowing both the number of atoms per unit cell and the overall mass, Avogadro's Number allows us to work backwards to understand the atomic composition at a larger scale.
Using Avogadro's Number provides a comprehensive view of how an immense number of atoms build up the macroscopic properties of materials we measure in grams.

Molar Mass Calculation

Molar mass links the atomic scale to comprehensive daily measurements, like grams. It tells us how much a mole of a substance weighs. The calculation involves using information derived from the mass and volume of the crystal's unit cell, as well as the number of atoms it contains. Given the formula:

• \( \text{Molar mass} = \frac{\text{mass of unit cell} \times N_{A}}{n} \)

Where \(N_{A}\) is Avogadro's Number, and \(n\) the number of atoms per unit cell (in an FCC structure, this is 4). By plugging in the known values such as mass of the unit cell and Avogadro's Number, we solve for the molar mass.
The molar mass is a bridge between the microscopic world of atoms and the macroscopic world of grams and liters, enabling chemists and students alike to precisely measure and predict the quantities and behaviors of substances in chemical reactions.