Question

What is the number of moles and total number of atoms respectively present in 5.6 cm\(^3\) of ammonia gas at STP?

• A. \( 2.50 \times 10^{-3} \) mol and \( 1.5 \times 10^{20} \) atoms
• B. \( 1.505 \) mol and \( 6.022 \times 10^{20} \) atoms
• C. \( 2.05 \) mol and \( 1.50 \times 10^{20} \) atoms
• D. \( 2.50 \times 10^{-4} \) mol and \( 6.022 \times 10^{20} \) atoms

Hint:

At STP, 1 mole of gas occupies 22.4 L. Use this to calculate the moles and number of atoms for any given volume of gas.

Answer 1

The Correct Option is D

Step 1: Using the ideal gas law.
At STP, 1 mole of any ideal gas occupies 22.4 L. First, we convert 5.6 cm\(^3\) to liters: \[ 5.6 \, \text{cm}^3 = 5.6 \times 10^{-3} \, \text{L} \] The number of moles \( n \) is: \[ n = \frac{V}{22.4} = \frac{5.6 \times 10^{-3}}{22.4} = 2.50 \times 10^{-4} \, \text{mol} \]
Step 2: Total number of atoms.
Since each molecule of ammonia (NH\(_3\)) contains 4 atoms (1 nitrogen and 3 hydrogens), the total number of atoms is: \[ \text{Atoms} = n \times 4 \times N_A = 2.50 \times 10^{-4} \times 4 \times 6.022 \times 10^{23} = 6.022 \times 10^{20} \, \text{atoms} \]
Step 3: Conclusion.
The correct answer is (D) \( 2.50 \times 10^{-4} \) mol and \( 6.022 \times 10^{20} \) atoms.