Problem 102 - A Complete Resource Book in Chemistry for JEE Main

Question

Which has maximum number of molecules? (a) \(7 \mathrm{~g} \mathrm{~N}_{2}\) (b) \(2 \mathrm{~g} \mathrm{H}_{2}\) (c) \(18 \mathrm{~g} \mathrm{NO}_{2}\) (d) \(16 \mathrm{~g} \mathrm{O}_{2}\)

Step-by-Step Solution

Answer

2 g of \( \mathrm{H}_2 \) contains the maximum number of molecules.

Step 1: Understanding the Problem

We need to determine which of the given substances contains the maximum number of molecules. We do this by calculating the number of moles for each substance, then using Avogadro's number to find the number of molecules.

Step 2: Calculate Moles for Each Substance

First, find the molar mass of each substance and then use it to calculate the moles:- For \( \mathrm{N}_2 \), the molar mass is \(2 \times 14 = 28 \; \mathrm{g/mol}\). Thus, \(7 \; \mathrm{g} \; \mathrm{N}_2\) is \(\frac{7}{28} = 0.25 \; \text{moles}\).- For \( \mathrm{H}_2 \), the molar mass is \(2 \times 1 = 2 \; \mathrm{g/mol}\). Thus, \(2 \; \mathrm{g} \; \mathrm{H}_2\) is \(\frac{2}{2} = 1 \; \text{mole}\).- For \( \mathrm{NO}_2 \), the molar mass is \(14 + 2 \times 16 = 46 \; \mathrm{g/mol}\). Thus, \(18 \; \mathrm{g} \; \mathrm{NO}_2\) is \(\frac{18}{46} \approx 0.391 \; \text{moles}\).- For \( \mathrm{O}_2 \), the molar mass is \(2 \times 16 = 32 \; \mathrm{g/mol}\). Thus, \(16 \; \mathrm{g} \; \mathrm{O}_2\) is \(\frac{16}{32} = 0.5 \; \text{moles}\).

Step 3: Finding Number of Molecules

Using Avogadro’s number \(6.022 \times 10^{23}\), calculate the number of molecules for each amount of substance:- \( \mathrm{N}_2 \): \(0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}\) molecules.- \( \mathrm{H}_2 \): \(1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}\) molecules.- \( \mathrm{NO}_2 \): \(0.391 \times 6.022 \times 10^{23} \approx 2.353 \times 10^{23}\) molecules.- \( \mathrm{O}_2 \): \(0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\) molecules.

Step 4: Comparison and Conclusion

Compare the number of molecules calculated for each substance. The number of molecules for \( \mathrm{H}_2 \) is \(6.022 \times 10^{23}\), which is the maximum compared to the others. Therefore, \(2 \; \mathrm{g} \; \mathrm{H}_2\) has the maximum number of molecules.

Key Concepts

Avogadro's Number

Avogadro's Number is a fundamental concept in chemistry that allows us to understand how many entities, like atoms or molecules, are present in a mole of a substance. Named after the Italian scientist Amedeo Avogadro, the number is defined as \[6.022 imes 10^{23} \]This seemingly enormous number represents the quantity of molecules or atoms in one mole. It's essential because atoms and molecules are extremely small, making them impossible to count individually in usual quantities.
With Avogadro's Number, chemists can convert between the amount of material (in moles) and the number of molecules or atoms that make up that material. For any element or compound, multiplying the number of moles by Avogadro's Number gives us the total number of molecules or atoms present. For example, 0.5 moles of oxygen molecules (\( \text{O}_2 \)) means you have \[0.5 imes 6.022 imes 10^{23} = 3.011 imes 10^{23} \]molecules of oxygen.

Molar Mass Calculation

Molar mass is a critical factor when determining the number of moles in a given sample. It is the mass of one mole of a substance and is expressed in grams per mole (\( \text{g/mol} \)). The molar mass of a molecule can be calculated by summing the atomic masses of all the atoms in the molecule, which are easily available on the periodic table.For instance:

• Nitrogen (\( \text{N}_2 \)): The molar mass is calculated as \( 2 \times 14 \; \text{g/mol} = 28 \; \text{g/mol} \) since the atomic mass of nitrogen is approximately 14.
• Hydrogen (\( \text{H}_2 \)): Its molar mass is \( 2 \times 1 \; \text{g/mol} = 2 \; \text{g/mol} \), with hydrogen's atomic mass being 1.
• Nitrogen Dioxide (\( \text{NO}_2 \)): Combine the atomic masses of nitrogen and oxygen: \( 14 + 2 \times 16 \; \text{g/mol} = 46 \; \text{g/mol} \).
• Oxygen (\( \text{O}_2 \)): Here, the molar mass is \( 2 \times 16 \; \text{g/mol} = 32 \; \text{g/mol} \).

After finding the molar mass, calculating the number of moles in any given mass of substance involves dividing the mass of the sample by the molar mass. This helps to compare the amounts of different substances quantitatively, as seen in the given problem. For example, for \( 16 \; \text{g} \) of \( \text{O}_2 \), the moles would be calculated as:\[\frac{16}{32} = 0.5 \; \text{moles}\]

Molecular Comparison

To find out which substance contains the maximum number of molecules, we first determine the number of moles of each substance and then use Avogadro’s Number to calculate the total number of molecules for each case.
Molecular comparison involves evaluating which sample has the highest count based on the molecule's count, not its mass. By taking each calculated mole:

• \(\text{N}_2\): 0.25 moles result in \( 0.25 \times 6.022 \times 10^{23} \approx 1.5055 \times 10^{23} \) molecules.
• \(\text{H}_2\): With 1 mole leading to \( 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23} \) molecules.
• \(\text{NO}_2\): Approximately 0.391 moles yield \( 0.391 \times 6.022 \times 10^{23} \approx 2.353 \times 10^{23} \) molecules.
• \(\text{O}_2\): Here, 0.5 moles will have \( 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \) molecules.

Through this comparison, \( 2 \; \text{g} \) of \( \text{H}_2 \), which contains \( 6.022 \times 10^{23} \) molecules, has the maximum number of molecules among the options given. This indicates that despite its small mass, \( \text{H}_2 \) contains the highest number of molecules because it has the smallest molecular weight, allowing more molecules per gram.