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 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.