Problem 70 - A Complete Resource Book in Chemistry for JEE Main

Question

\(1 \mathrm{cc} \mathrm{N}_{2} \mathrm{O}\) at NTP contains (a) \(\frac{1.32}{224} \times 10^{23}\) electrons (b) \(\frac{6.02}{22400} \times 10^{23}\) molecules (c) \(\frac{1.8}{224} \times 10^{22}\) atoms (d) all of these

Step-by-Step Solution

Answer

The correct answer is (d) all of these.

Step 1: Calculate Number of Molecules

At NTP (Normal Temperature and Pressure), 1 mole of a gas occupies 22,400 cc (cubic centimeters). To find the number of molecules in 1 cc of \( N_2 O \), we apply the formula: \[ \text{Number of molecules} = \left( \frac{6.02 \times 10^{23}}{22,400} \right) \times 1 \] Simplifying this, we get: \[ \frac{6.02}{22,400} \times 10^{23} \] Thus, the number of molecules in 1 cc of \( N_2 O \) is \( \frac{6.02}{22400} \times 10^{23} \).

Step 2: Calculate Number of Atoms

Each molecule of \( N_2 O \) contains 3 atoms (2 Nitrogen and 1 Oxygen atom). Therefore, the number of atoms in \( \frac{6.02}{22400} \times 10^{23} \) molecules is: \[ \text{Number of atoms} = \frac{6.02 \times 3}{22400} \times 10^{23} = \frac{18.06}{22400} \times 10^{23} \] Simplifying: \[ \frac{18.06}{22400} \times 10^{23} \approx \frac{1.8}{224} \times 10^{22} \] So, the number of atoms is \( \frac{1.8}{224} \times 10^{22} \).

Step 3: Calculate Number of Electrons

The molecular formula \( N_2 O \) has 2 Nitrogen atoms and 1 Oxygen atom. Each nitrogen atom contributes 7 electrons and each oxygen atom contributes 8 electrons, leading to a total of \( 2 \times 7 + 8 = 22 \) electrons per molecule. Therefore, the number of electrons is: \[ \text{Number of electrons} = \frac{6.02 \times 22}{22400} \times 10^{23} \] Simplifying: \[ \frac{132.44}{22400} \times 10^{23} \approx \frac{1.32}{224} \times 10^{23} \] Thus, the number of electrons is \( \frac{1.32}{224} \times 10^{23} \).

Step 4: Analyze Options

The calculations for the number of molecules, atoms, and electrons match with options (b), (c), and (a) respectively. Therefore, all given options are correct.

Key Concepts

Number of Molecules

Understanding the concept of counting molecules in a given volume at standard conditions can be crucial in chemistry. At Normal Temperature and Pressure (NTP), a mole of any gas occupies exactly 22,400 cubic centimeters (cc). This universal principle can be used to calculate the number of molecules within any smaller volume. For instance, in the exercise provided, you are asked to find the number of molecules in 1 cc of nitrous oxide \( N_2O \). By proportionally scaling the 1 mole benchmark, you calculate:

• Start with Avogadro's number, 6.02 x 10^{23} molecules per mole.
• Divide by the standard volume, 22,400 cc, to determine molecules per cc.
• Thus, in 1 cc of \( N_2O \), calculate \( \frac{6.02 \times 10^{23}}{22,400} \), which results in \( \frac{6.02}{22400} \times 10^{23} \) molecules.

This calculation provides a foundational understanding of how gas molecules are measured more intricately at smaller scales.

Number of Atoms

When it comes to determining the number of atoms from a given number of molecules, the molecular structure plays a key role. For \( N_2O \), which contains 3 atoms per molecule (2 nitrogen and 1 oxygen), it's straightforward to extrapolate the total atomic count. By using the number of molecules determined in the previous step, the process is as follows:

• Each molecule of \( N_2O \) contributes 3 atoms: two from nitrogen and one from oxygen.
• Multiply the number of molecules, \( \frac{6.02}{22400} \times 10^{23} \), by 3.
• This yields \( \frac{18.06}{22400} \times 10^{23} \), which simplifies approximately to \( \frac{1.8}{224} \times 10^{22} \) atoms.

Understanding the relationship between molecules and atoms is essential for comprehending the chemical complexity of compounds.

Number of Electrons

Counting the number of electrons in a given volume of gas involves knowing both the molecular composition and the atomic structure. In our example of \( N_2O \):

• The molecular formula reveals 2 nitrogen atoms (each with 7 electrons) and 1 oxygen atom (with 8 electrons).
• This totals to 22 electrons per molecule.
• Using the number of molecules previously determined, \( \frac{6.02}{22400} \times 10^{23} \), simply multiply by 22 for the electron count.
• This results in \( \frac{132.44}{22400} \times 10^{23} \), which simplifies to \( \frac{1.32}{224} \times 10^{23} \) electrons.

Through these calculations, we seamlessly bridge molecular composition to electron count, providing insights into atomic interactions and chemical behavior.