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

Concept:

Number of moles $= \dfrac{\text{Number of molecules}}{N_A}$
Total initial moles $=$ moles remaining $+$ moles removed
Mass $=$ moles $\times$ molar mass
Step 1: Moles of CO$_2$ remaining: \[ n_{\text{left}} = 2.8 \times 10^{-3}\,\text{mol} \]
Step 2: Moles of CO$_2$ removed: \[ n_{\text{removed}} = \frac{10^{21}}{6.02 \times 10^{23}} \] \[ n_{\text{removed}} \approx 1.66 \times 10^{-3}\,\text{mol} \]
Step 3: Initial number of moles of CO$_2$: \[ n_{\text{initial}} = n_{\text{left}} + n_{\text{removed}} \] \[ n_{\text{initial}} = 2.8 \times 10^{-3} + 1.66 \times 10^{-3} \] \[ n_{\text{initial}} = 4.46 \times 10^{-3}\,\text{mol} \]
Step 4: Molar mass of CO$_2$: \[ M = 12 + 2 \times 16 = 44\,\text{g/mol} \]
Step 5: Initial mass of CO$_2$: \[ m = n_{\text{initial}} \times M = 4.46 \times 10^{-3} \times 44 \] \[ m = 0.19624\,\text{g} = 196.2\,\text{mg} \]