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

Given:
\( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \) (Avogadro's number)
\( \text{mol}_{\text{initial}} = \frac{x \times 10^{-3}}{44} \)
\( \text{mol}_{\text{removal}} = \frac{10^{21}}{6.02 \times 10^{23}} \)
Now, the total moles left can be written as: \[ \text{mol}_{\text{left}} = \text{mol}_{\text{initial}} - \text{mol}_{\text{removal}} \] Substitute the given values: \[ 2.8 \times 10^{-3} = \frac{x \times 10^{-3}}{44} - \frac{10^{21}}{6.02 \times 10^{23}} \] Simplifying the equation: \[ 2.8 \times 10^{-3} = \frac{x \times 10^{-3}}{44} - 1.66 \times 10^{-3} \] Now, solve for \( x \): \[ x \times 10^{-3} = (2.8 + 1.66) \times 10^{-3} \times 44 \] \[ x = 196.2 \, \text{mg} \] Thus, the mass of \( \text{CO}_2 \) initially taken is \( 196.2 \, \text{mg} \).