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

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Question

(A) The molar mass of $CO_2 = 44 \ g/mol$. Given mass $= 200 \ mg = 0.2 \ g$. Initial moles of $CO_2 = \frac{0.2 \ g}{44 \ g/mol} \approx 0.004545 \ m

Vedclass · Accepted Answer

$2.84 	imes 10^{-3}$

Vedclass · Answer

(A) The molar mass of $CO_2 = 44 \ g/mol$. Given mass $= 200 \ mg = 0.2 \ g$. Initial moles of $CO_2 = \frac{0.2 \ g}{44 \ g/mol} \approx 0.004545 \ mol$. 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 $= 2.737 	imes 10^{21} - 1.0 	imes 10^{21} = 1.737 	imes 10^{21}$ molecules. Remaining moles $= \frac{1.737 	imes 10^{21}}{6.022 	imes 10^{23}} \approx 0.288 	imes 10^{-2} \ mol = 2.88 	imes 10^{-3} \ mol$. Rounding to the nearest provided option,the answer is $2.84 	imes 10^{-3}$.

If $10^{21}$ molecules are removed from $200 \ mg$ of $CO_2$,how many moles of $CO_2$ will remain?

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