Question

${10}^{21}$ molecules are removed from $200mg$ of $C{O}_2$. The moles of $C{O}_2$ left are :

• A. $2.88\times {10}^{-3}$
• B. $28.8\times {10}^{-3}$
• C. $288\times {10}^{-3}$
• D. $28.8\times {10}^3$

Answer 1

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

First , to determine the number of moles of $C{O}_2$ in $200g$.

As formula is :

No. of moles $=\frac{Wt.ing}{Mol.wt}$

No. of moles in $200mg=\frac{200}{1000\times 44}=4.54\times {10}^{-3}moles$

So now, No. of moles in ${10}^{21}$ molecules$=\frac{{10}^{21}}{6.02\times {10}^{23}}=1.67\times {10}^{-3}moles$

Now to get the number of moles left when ${10}^{21}$ moles are removed from the $200g$ or $4.54\times {10}^{-3}moles$, following calculation is done:

No. of moles left $=(4.54-1.67)\times {10}^{-3}=2.88\times {10}^{-3}$ mol