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

To determine the initial mass of the \( \text{CO}_2 \) sample, we perform the following steps: 

1. First, determine the number of molecules initially present in the \( x \) mg sample. Since \( 10^{21} \) molecules are removed, and \( 2.8 \times 10^{-3} \) moles of \( \text{CO}_2 \) remain, calculate the number of molecules in \( 2.8 \times 10^{-3} \) moles: \(N = n \times N_A = 2.8 \times 10^{-3} \times 6.02 \times 10^{23}\). \(N = 1.6856 \times 10^{21} \text{ molecules}\).
2. Add the removed molecules to find the original number of molecules: \(N_{total} = 1.6856 \times 10^{21} + 10^{21} = 2.6856 \times 10^{21} \text{ molecules}\).
3. Convert the total number of molecules into 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. Knowing the molar mass of \( \text{CO}_2 \) is approximately 44 g/mol, calculate the mass initially present: \(m = n \times \text{Molar Mass} = 4.46 \times 10^{-3} \times 44 \times 10^3 \, \text{mg}\). \(m \approx 196.2 \, \text{mg}\).

Thus, the initial mass of the \( \text{CO}_2 \) sample is 196.2 mg. This corresponds to the correct answer.

Answer 2

To find the initial mass of \( \text{CO}_2 \), first calculate the number of moles corresponding to \( 10^{21} \) molecules. 

Using Avogadro's number \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \):

\[ \frac{10^{21}}{6.02 \times 10^{23}} = 1.66 \times 10^{-3} \, \text{mol} \]

The initial moles of \( \text{CO}_2 \) are:

\[ 2.8 \times 10^{-3} + 1.66 \times 10^{-3} = 4.46 \times 10^{-3} \, \text{mol} \]

The molar mass of \( \text{CO}_2 \) is approximately \( 44 \, \text{g/mol} \). 

Hence, the mass of \( \text{CO}_2 \) is:

\[ 4.46 \times 10^{-3} \, \text{mol} \times 44 \, \text{g/mol} = 196.24 \, \text{mg} \]

Thus, the initial mass of \( \text{CO}_2 \) is 196.2 mg.