Question

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

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

Hint:

When particles are removed, always convert them into moles first and add to the remaining moles to get the initial amount.

Answer 1

The Correct Option is C

1. First, we need to determine the number of moles that correspond to the $10^{21}$ molecules of CO₂ that were removed. Given that Avogadro's number \(N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}\), we can use the formula for the number of moles: 
   \(n = \frac{N}{N_A}\) 
   where \(n\) is the number of moles and \(N\) is the number of molecules. 
   Substituting the given values: 
   \(n = \frac{10^{21}}{6.02 \times 10^{23}} = 1.66 \times 10^{-3}\,\text{mol}\)
2. According to the problem, after removing these molecules, \(2.8 \times 10^{-3}\) moles of CO₂ remain. Therefore, the initial number of moles of CO₂ was: 
   \(n_{\text{initial}} = n + 2.8 \times 10^{-3} = 1.66 \times 10^{-3} + 2.8 \times 10^{-3} = 4.46 \times 10^{-3}\,\text{mol}\)
3. The molar mass of CO₂ is the sum of the masses of one carbon atom and two oxygen atoms. Carbon has an atomic mass of 12 u, and oxygen has an atomic mass of 16 u, therefore the molar mass of CO₂ is: 
   \(12 + 2 \times 16 = 44\, \text{g/mol}\)
4. Using the formula for mass: 
   \(m = n_{\text{initial}} \times \text{Molar Mass}\) 
   we find the initial mass of CO₂: 
   \(m = 4.46 \times 10^{-3} \times 44 = 0.19624\, \text{g} = 196.2\, \text{mg}\)

Thus, the initial mass of CO₂ that was taken is 196.2 mg. Therefore, the correct answer is 196.2 mg.