Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

If molar mass of \(\mathrm{AB}\) is \(95 \mathrm{~g} \mathrm{~mol}^{-1}\) and \(\mathrm{a}\) is edge length, then the density of crystal structure is (a) \(\frac{4 \times 95}{a^{3} \times N_{A}}\) (b) \(\frac{4 \mathrm{~N}_{\mathrm{A}}}{\mathrm{a}^{3} \times 95}\) (c) \(\frac{2 \times 95}{a^{3} \times N_{4}}\) (d) \(\frac{2 \mathrm{~N}_{\mathrm{A}}}{\mathrm{a}^{3} \times 95}\)

Step-by-Step Solution

Answer

The correct answer is (a) \( \frac{4 \times 95}{a^{3} \times N_{A}} \).

Step 1: Understanding Density Formula for Crystals

Density \( (\rho) \) of a crystal structure is given by the formula \( \rho = \frac{Z \times M}{a^3 \times N_A} \), where \( Z \) is the number of formula units per unit cell, \( M \) is the molar mass of the compound, \( a \) is the edge length of the cube, and \( N_A \) is Avogadro's number, approximately \( 6.022 \times 10^{23} \text{ mol}^{-1} \).

Step 2: Identify the Values in the Problem

From the provided options and given data, we know the molar mass \( M \) is \( 95 \text{ g mol}^{-1} \). The crystalline structure suggests it might be a type of cubic cell. In simple cubic structures, typically \( Z = 1 \), in face-centered cubic \( Z = 4 \), and in body-centered cubic \( Z = 2 \). Since the problem uses \( Z = 4 \) in the density formula, assume the unit cell is face-centered cubic.

Step 3: Substitute Values and Form the Density Equation

Using the density formula for a face-centered cubic structure: \( \rho = \frac{4 \times 95}{a^3 \times N_A} \). Here, \( Z = 4 \), \( M = 95 \text{ g mol}^{-1} \), \( a \) is the edge length, and \( N_A \) is Avogadro's number.

Step 4: Compare Formed Equation with Given Options

The equation formed, \( \rho = \frac{4 \times 95}{a^3 \times N_A} \), matches closely with option (a). Review the other options to ensure no mathematical similarity, eventually verifying that option (a) is the correct match.

Key Concepts

Face-Centered Cubic Cell

The face-centered cubic (FCC) cell is a prevalent crystal structure in chemistry. Imagine a cube where atoms are located at each corner and in the center of each face of the cube. This arrangement provides optimal packing because each atom at the face shares its position with two adjacent cells.
In nature, elements like aluminum, copper, and gold often crystallize in an FCC structure because it is dense and these metals closely pack atoms efficiently. This dense arrangement allows them to have strong metallic bonds.
In the FCC cell, there are 8 atoms at the corners (each shared by 8 adjacent cells) and 6 center face atoms (each shared by 2 cells), contributing effectively to 4 whole atoms per unit cell. This is why the variable \( Z \), which denotes the number of atoms per unit cell, is 4 in an FCC structure.
Knowing the structure type, such as FCC, helps us accurately calculate important properties such as density using relationships that account for how atoms are arranged in space. Understanding this grid-like foundation aids in mastering crystallography.

Avogadro's Number

Avogadro's number, approximately \( 6.022 \times 10^{23} \text{ mol}^{-1} \), is a fundamental constant in chemistry that signifies the number of atoms, ions, or molecules in one mole of a substance. This vast quantity allows chemists to translate between the atomic scale and quantities we can measure in the lab.
For instance, when calculating the density of a crystal, Avogadro's number helps determine how many fundamental particles make up the substance at a provided scale. This is crucial because chemistry often deals with atomic and molecular entities that are too small to count individually. Hence, Avogadro's number becomes a connective bridge, turning atomic mass into measurable grams.
This number finds its use in a wide range of calculations including, but not limited to, molar volume and conversion between moles and molecules. Recognizing the sheer vastness of Avogadro's number underscores the small scale at which atoms and molecules operate, making it an essential concept for students of chemistry.

Molar Mass Calculation

Molar mass is the weight, in grams, of one mole of a given substance's atoms. It acts as a form of measurement connecting the mass of individual atoms or molecules to a convenient bulk scale, which allows chemists to create reactions and solutions according to precise molecular needs.
Calculating molar mass involves summing the atomic masses of all atoms in a molecular formula. For example, if we consider the compound AB, and its molar mass is given as \( 95 \text{ g mol}^{-1} \), it means that one mole of AB weighs 95 grams.
This metric is also essential in the context of density calculations for crystals, where the molar mass helps establish how much a specific volume of a crystalline material should weigh under ideal conditions relating to its organized structure.
Molar mass bridges the molecular world with our tangible one, making it indispensable for stoichiometry and quantitative chemistry analyses. Understanding how to perform and apply molar mass calculations is therefore a foundational skill in any chemistry curriculum.