Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

The density of solid argon is \(1.65 \mathrm{~g} / \mathrm{mL}\) at \(-233^{\circ} \mathrm{C}\). If the argon atom is assumed to be sphere of radius \(1.54 \times\) \(10^{-8} \mathrm{~cm}\), what percentage of solid argon is apprarently empty space? (Atomic wt of \(\mathrm{Ar}=40\) ) (a) \(32 \%\) (b) \(52 \%\) (c) \(62 \%\) (d) \(72 \%\)

Step-by-Step Solution

Answer

The percentage of empty space is approximately 62%, which corresponds to option (c).

Step 1: Calculate the Volume of a Single Argon Atom

The volume of an atom assuming it to be a sphere can be calculated using the formula \( V = \frac{4}{3}\pi r^3 \). Here, \( r = 1.54 \times 10^{-8} \) cm.Calculate the volume:\(V = \frac{4}{3} \pi (1.54 \times 10^{-8})^3 \\approx 1.53 \times 10^{-23} \text{ cm}^3\)

Step 2: Calculate the Mass of a Single Argon Atom

First, calculate the number of moles in 1 gram of argon using the atomic weight of argon (40):\(\frac{1}{40} \text{ mol}\)Using Avogadro's number \(6.022 \times 10^{23} \text{ atoms/mol}\), find the mass of one atom:\(\text{Mass of one atom} = \frac{1 \, \text{g}}{40} \times \frac{1}{6.022 \times 10^{23}} \approx 6.64 \times 10^{-23} \text{ g}\)

Step 3: Density Relation and Volume Occupied Calculation

The density of solid argon is given by \( \frac{\text{mass}}{\text{volume}} = 1.65 \, \text{g/mL} \), which equals \(1.65 \, \text{g/cm}^3\).Calculate volume occupied by one argon atom in the crystal:\(\text{Volume} = \frac{\text{Mass of one atom}}{\text{Density}} = \frac{6.64 \times 10^{-23} \text{ g}}{1.65 \, \text{g/cm}^3} \approx 4.02 \times 10^{-23} \text{ cm}^3\)

Step 4: Calculate Percentage of Empty Space

The percentage of empty space is the difference between the volume of one argon atom and the volume it occupies in the solid, divided by the volume it occupies, multiplied by 100.Compute the empty space:\(\text{empty space %} = \left(1 - \frac{\text{Volume of the atom}}{\text{Volume it occupies}}\right) \times 100 \= \left(1 - \frac{1.53 \times 10^{-23}}{4.02 \times 10^{-23}}\right) \times 100 \approx 62\%\)

Step 5: Conclusion: Answer Selection

Based on the calculation, the percentage of solid argon that is empty space is approximately \(62\%\).Thus, the correct answer is (c) \(62\%\).

Key Concepts

Volume of an Atom

Atoms are the basic building blocks of matter. Each type of atom can be visualized as a small sphere, and like any sphere, the volume can be computed using the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \).
In this context, the radius \(r\) of the argon atom is given as \(1.54 \times 10^{-8} \text{ cm}\). By substituting this value into the formula, we can find the volume of a single argon atom is approximately \(1.53 \times 10^{-23} \text{ cm}^3\).
This volume is extremely small, which is why chemical and physical changes seem continuous and are not observed at the atomic level.

Atomic Weight of Argon

The atomic weight of an element refers to the average mass of atoms of an element, calculated using the relative abundance of isotopes in a naturally-occurring element.
For argon, the atomic weight is about 40. This means that in one mole (a standard scientific unit for measuring large quantities of very small entities), there are approximately 40 grams of argon.

• Since the atomic weight provides the mass of one mole, it is pivotal for determining the mass of single atoms when using Avogadro's number (for example, when calculating the mass of a single argon atom).
• Knowing the atomic weight helps scientists and students relate the scale of individual atoms to macroscopic quantities.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry, which is \(6.022 \times 10^{23}\). It defines the quantity of atoms, molecules, or ions in one mole of substance.
Using this number, the mass of a single argon atom can be calculated from its atomic weight. For instance, knowing that 40 grams of argon correspond to one mole (or \(6.022 \times 10^{23}\) atoms), we can derive the mass of a single atom through a straightforward division, yielding approximately \(6.64 \times 10^{-23} \text{ g}\).

• Avogadro's number acts as a bridge between the atomic scale and the macroscopic scale.
• It is crucial for calculating quantities for chemical reactions and material properties where individual atoms or molecules are involved.

Percentage of Empty Space

In solid materials like argon at very low temperatures, individual atoms are arranged into a lattice structure. However, there is often a significant amount of empty space between these atoms.
To calculate the percentage of empty space within solid argon, you'll first need the volume occupied by one atom and its actual density-derived volume in the crystal.
This is found with the equation: \( \text{empty space %} = \left(1 - \frac{\text{Volume of the atom}}{\text{Volume it occupies}}\right) \times 100 \).
In our argon example, it turns out that approximately \(62\%\) of the space is empty. This highlights the efficient but not perfect packing of atoms in solid argon.
This calculation not only illuminates the inner structure of solids but also assists in understanding molecular interactions and properties of substances at different phases.