Question

2.8 \( \times 10^{-3} \) mol of \( \text{CO}_2 \) is left after removing \( 10^{21} \) molecules from its ‘\( x \)’ mg sample. The mass of \( \text{CO}_2 \) taken initially is: Given: \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)

• A. 196.2 mg
• B. 98.3 mg
• C. 150.4 mg
• D. 48.2 mg

Hint:

In problems involving moles and mass, always use the relationship \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Ensure correct conversion between molecules and moles using Avogadro's number.

Answer 1

The Correct Option is A

The initial mass of the \( \text{CO}_2 \) sample is determined through the following procedure:

1. Calculate the number of molecules in the remaining \( 2.8 \times 10^{-3} \) moles of \( \text{CO}_2 \) using the Avogadro constant (\( N_A \)): \(N = n \times N_A = 2.8 \times 10^{-3} \times 6.02 \times 10^{23} = 1.6856 \times 10^{21} \) molecules.
2. Determine the total initial number of molecules by adding the removed molecules: \(N_{total} = 1.6856 \times 10^{21} + 10^{21} = 2.6856 \times 10^{21} \) molecules.
3. Convert the total number of molecules back to moles: \(n = \frac{N_{total}}{N_A} = \frac{2.6856 \times 10^{21}}{6.02 \times 10^{23}} \approx 4.46 \times 10^{-3} \, \text{mol}\).
4. Calculate the initial mass of the \( \text{CO}_2 \) sample using its molar mass (approximately 44 g/mol): \(m = n \times \text{Molar Mass} = 4.46 \times 10^{-3} \times 44 \times 10^3 \, \text{mg} \approx 196.2 \, \text{mg}\).

Consequently, the initial mass of the \( \text{CO}_2 \) sample was 196.2 mg.