Question

From $220\text{mg}$ of $C{O}_2$, ${10}^{21}$ molecules are removed. How many moles of $C{O}_2$ are left?
Take $N_A=6\times {10}^{23}$.

• A. $3.33\times {10}^{-3}\text{mol}$
• B. $7.2\times {10}^{-2}\text{mol}$
• C. $3.33\times {10}^{-1}\text{mol}$
• D. $4.2\text{mol}$

Answer 1

The correct option is A $3.33\times {10}^{-3}\text{mol}$

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 $=\frac{44\times {10}^{21}}{6\times {10}^{23}}=\frac{220}{3}\text{mg of}C{O}_2$
Amount of $C{O}_2$ left $=\text{Initial amount}-\text{Removed amount}=(220-\frac{220}{3})\text{mg}=\frac{440}{3}\text{mg}$
Number of moles $C{O}_2$ left $=\frac{\text{amount of}C{O}_2\text{left (g)}}{\text{molar mass (g/mol)}}=\frac{440\times {10}^{-3}}{3\times 44}=3.33\times {10}^{-3}\text{mol}$