2.8 times 10^{-3} mol of CO_2 is left after | vedclass

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Question

(A) The molar mass of $CO_2$ is $44 \ g/mol$. Initial moles of $CO_2$ = $\frac{x 	imes 10^{-3} \ g}{44 \ g/mol} = \frac{x 	imes 10^{-3}}{44} \ mol$.

Vedclass · Accepted Answer

$196.2$

Vedclass · Answer

(A) The molar mass of $CO_2$ is $44 \ g/mol$. Initial moles of $CO_2$ = $\frac{x 	imes 10^{-3} \ g}{44 \ g/mol} = \frac{x 	imes 10^{-3}}{44} \ mol$. Moles of $CO_2$ removed = $\frac{10^{21}}{6.02 	imes 10^{23}} \approx 1.66 	imes 10^{-3} \ mol$. Moles left = $	ext{Initial moles} - 	ext{Removed moles}$. $2.8 	imes 10^{-3} = \frac{x 	imes 10^{-3}}{44} - 1.66 	imes 10^{-3}$. $2.8 	imes 10^{-3} + 1.66 	imes 10^{-3} = \frac{x 	imes 10^{-3}}{44}$. $4.46 	imes 10^{-3} = \frac{x 	imes 10^{-3}}{44}$. $x = 4.46 	imes 44 = 196.24 \ mg$.

$2.8 \times 10^{-3} \ mol$ of $CO_2$ is left after removing $10^{21}$ molecules from its '$x$' $mg$ sample. The mass of $CO_2$ taken initially is. Given: $N_A = 6.02 \times 10^{23} \ mol^{-1}$. (in $mg$)

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