If 10^{21} molecules are removed from 200 mg | vedclass

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Question

(A) Initial mass of $CO_2 = 200 \ mg = 0.2 \ g$.Molar mass of $CO_2 = 44 \ g/mol$.Initial number of moles of $CO_2 = \frac{0.2 \ g}{44 \ g/mol} = 0.00

Vedclass · Accepted Answer

$2.85$

Vedclass · Answer

(A) Initial mass of $CO_2 = 200 \ mg = 0.2 \ g$.Molar mass of $CO_2 = 44 \ g/mol$.Initial number of moles of $CO_2 = \frac{0.2 \ g}{44 \ g/mol} = 0.004545 \ mol$.Initial number of molecules $= 0.004545 	imes 6.022 	imes 10^{23} \approx 2.737 	imes 10^{21}$ molecules.Number of molecules removed $= 10^{21}$.Remaining molecules $= 2.737 	imes 10^{21} - 1 	imes 10^{21} = 1.737 	imes 10^{21}$ molecules.Remaining moles $= \frac{1.737 	imes 10^{21}}{6.022 	imes 10^{23}} \approx 0.288 	imes 10^{-2} = 2.88 	imes 10^{-3} \ mol$.Note: Using $6 	imes 10^{23}$ as Avogadro's number gives $2.72 	imes 10^{21}$ molecules,leading to $1.72 	imes 10^{21}$ remaining,which results in $\frac{1.72}{6} 	imes 10^{-2} \approx 0.286 	imes 10^{-2} = 2.86 	imes 10^{-3} \ mol$. The closest option is $2.85$.

If $10^{21}$ molecules are removed from $200 \ mg$ of $CO_2$,then the number of moles of $CO_2$ left are .......... $\times 10^{-3}$.

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