Question

Find the number of atoms present in \(11.2\ \text{L}\) of nitrogen gas at STP.

• A. \(3.01 \times 10^{23}\)
• B. \(6.02 \times 10^{23}\)
• C. \(12.04 \times 10^{23}\)
• D. \(1.204 \times 10^{23}\)

Hint:

Useful STP facts:
• \(22.4\ \text{L}\) gas at STP \(= 1\) mole
• \(1\) mole \(= 6.02 \times 10^{23}\) molecules
• For diatomic gases like \(H_2, N_2, O_2, Cl_2\): \[ \text{Atoms} = 2 \times \text{Number of molecules} \]

Answer 1

The Correct Option is B

Concept: At STP (Standard Temperature and Pressure), \[ 1 \text{ mole of any gas occupies } 22.4\ \text{L} \] Also, \[ 1 \text{ mole contains } 6.02 \times 10^{23} \text{ molecules} \] Nitrogen gas exists as a diatomic molecule: \[ N_2 \] Hence, each molecule of nitrogen contains: \[ 2 \text{ nitrogen atoms} \]

Step 1: Calculating the number of moles of nitrogen gas.
Given volume of nitrogen gas: \[ V = 11.2\ \text{L} \] At STP: \[ 22.4\ \text{L} = 1 \text{ mole} \] Therefore, \[ \text{Number of moles} = \frac{11.2}{22.4} \] \[ = 0.5 \text{ mole} \]

Step 2: Calculating the number of nitrogen molecules.
We know: \[ 1 \text{ mole} = 6.02 \times 10^{23} \text{ molecules} \] So, \[ 0.5 \text{ mole} = 0.5 \times 6.02 \times 10^{23} \] \[ = 3.01 \times 10^{23} \text{ molecules} \]

Step 3: Converting molecules into atoms.
Each molecule of nitrogen gas is: \[ N_2 \] Thus, each molecule contains \(2\) atoms. Therefore, \[ \text{Number of atoms} = 2 \times 3.01 \times 10^{23} \] \[ = 6.02 \times 10^{23} \]

Step 4: Final conclusion.
Hence, the number of atoms present in \(11.2\ \text{L}\) of nitrogen gas at STP is: \[ \boxed{6.02 \times 10^{23}} \] Therefore, the correct option is: \[ \boxed{(2)\ 6.02 \times 10^{23}} \]