Question

From $200\text{mg}$ of $C{O}_2$, ${10}^{21}$ molecules are removed. How many moles of $C{O}_2$ are left?

• A. $2.88\times {10}^{-3}\text{mol}$
• B. $7\times {10}^{-2}\text{mol}$
• C. $2.88\times {10}^{-1}\text{mol}$
• D. $2.8\text{mol}$

Answer 1

The correct option is A $2.88\times {10}^{-3}\text{mol}$

1 mole of $C{O}_2$ has a mass of $44\text{g}$ and contains $N_A$ number of molecules.
So, ${10}^{21}$ molecules will be present in $=\frac{44\times {10}^{21}}{6.023\times {10}^{23}}=0.0731\text{g of}C{O}_2=73.1\text{mg of}C{O}_2$
Amount of $C{O}_2$ remained $=\text{Initial amount}-\text{Removed amount}=(200-73.1)\text{mg}$
$=126.9\text{mg}$
Number of moles $C{O}_2$ left $=\frac{\text{amount of}C{O}_2\text{left (g)}}{\text{molar mass (g/mol)}}=\frac{126.9\times {10}^{-3}}{44}=2.88\times {10}^{-3}\text{moles}$