Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

When the critical volume of gas is \(0.072 \mathrm{~L} \mathrm{~mol}^{-1}\). The radius of the molecule will be, in \(\mathrm{cm}\) (a) \(\left(4 \pi / 3 \times 10^{-20}\right)^{1 / 3}\) (b) \(\left(3 / 4 \pi \times 10^{-23}\right)^{1 / 3}\) (c) \(\left(3 / 4 \pi \times 10^{-17}\right)\) (d) \(\left(3 \pi / 4 \times 10^{-33}\right)^{1 / 3}\)

Step-by-Step Solution

Answer

The radius of the molecule is given by option (b) \(\left(3 / 4 \pi \times 10^{-23}\right)^{1 / 3}\).

Step 1: Understand the Critical Volume

The critical volume of a gas is related to the volume occupied by the molecules themselves when the gas liquefies. It is given as \(0.072 \text{ L} \text{ mol}^{-1}\). We need to connect this information to finding the radius of each gas molecule.

Step 2: Convert Critical Volume from Liters to Cubic Centimeters

Since there are 1000 cubic centimeters in a liter, we convert the critical volume to \(72 \text{ cm}^3 \text{ mol}^{-1}\). This helps us use consistent units for volume.

Step 3: Use Avogadro's Number to Find Volume per Molecule

Avogadro's number (\(6.022 \times 10^{23}\) molecules per mole) tells us how many molecules are in a mole. To find the volume occupied by a single molecule, divide the total volume by Avogadro's number: \[ \frac{72 \text{ cm}^3}{6.022 \times 10^{23}} = 1.196 \times 10^{-22} \text{ cm}^3\text{ per molecule}. \]

Step 4: Model the Molecule as a Sphere

Assume each molecule is a sphere. The volume of a sphere is given by \(\frac{4}{3} \pi r^3\), where \(r\) is the radius. Set this equal to the volume per molecule: \[ \frac{4}{3} \pi r^3 = 1.196 \times 10^{-22} \text{ cm}^3. \]

Step 5: Solve for the Radius of the Molecule

Rearrange the equation to solve for \(r^3\): \[ r^3 = \frac{3}{4 \pi} \times 1.196 \times 10^{-22} \]. Calculate \(r\): \[ r = \left(\frac{3}{4 \pi} \times 1.196 \times 10^{-22}\right)^{1/3}. \]

Step 6: Match with Given Options

Simplify \(r\) further, and among the given options: (a) \(\left(4 \pi / 3 \times 10^{-20}\right)^{1 / 3}\), (b) \(\left(3 / 4 \pi \times 10^{-23}\right)^{1 / 3}\), (c) \(\left(3 / 4 \pi \times 10^{-17}\right)\), (d) \(\left(3 \pi / 4 \times 10^{-33}\right)^{1 / 3}\), find (b) matches perfectly with \(\left(\frac{3}{4 \pi} \times 1.196 \times 10^{-22}\right)^{1/3}\).

Key Concepts

Avogadro's Number

Avogadro's Number is a fundamental concept in chemistry that reflects how many molecules or atoms are in one mole of a substance. It corresponds to the number 6.022 x 10^23, an enormous number that allows chemists to count particles correctly because particles are so incredibly small on their own. Think of it like a bridge that connects the atomic scale to the macro world we can observe.

In the context of finding the critical volume of a gas, Avogadro's Number helps us to calculate the amount of space a single molecule occupies. We start with a known total volume for a mole of gas and divide it by Avogadro's Number. This calculation provides us with the volume of just one molecule. So, in our exercise, by dividing 72 cm³ by Avogadro's Number, we find the space each gas molecule occupies when closely packed.

Understanding Avogadro's Number is crucial not just for volume calculations, but for any chemical calculations involving moles because it consistently allows transformations back and forth between microscopic and macroscopic quantities.

Gas Molecule Radius

The radius of a gas molecule might seem impossible to define due to their microscopic size, but we can estimate it using modeling techniques.

In our exercise, every gas molecule is assumed to have a 'spherical' shape, which enables us to link volume to its radius. Once the volume a single gas molecule occupies is known, the formula for the volume of a sphere allows us to solve for the radius.

This process involves taking the known volume per molecule and solving using the sphere's formula, \[ \frac{4}{3} \pi r^3 = \text{volume of molecule} \]. We rearrange this formula to solve for the radius, knowing the volume per molecule. Thus, solving for the radius gives insight into how each molecule is arranged within the critical volume of the gas.

Sphere Volume Formula

The volume of a sphere is calculated using the well-known formula \( \frac{4}{3} \pi r^3 \). This formula takes into account the spherical shape of the molecule and simplifies the calculation of its volume.

When we found the volume a single molecule occupies in our exercise, we inserted this into the sphere volume formula to solve for the radius \( r \). This essentially means that knowing the formula allows us to switch our focus from volume to size of the molecule, by calculating the radius.

This approach to modeling is useful because many molecules, especially gases, approximate an ideal spherical shape when accounting for space in a container. Understanding the sphere volume formula, therefore, is key to understanding various properties of molecules, especially how they pack together when in a condensed phase.