$2.8 imes10^{-3}$ mol of $\mathrm{CO_2}$ is left after removing $10^{21}$ molecules from its ' $x$ ' mg sample. The mass of $\mathrm{CO_2}$ taken initially is Given: $N_A=6.02 imes10^{23}\ \mathrm{mol^{-1}}$

Question

$2.8	imes10^{-3}$ mol of $\mathrm{CO_2}$ is left after removing $10^{21}$ molecules from its ' $x$ ' mg sample. The mass of $\mathrm{CO_2}$ taken initially is Given: $N_A=6.02	imes10^{23}\ \mathrm{mol^{-1}}$

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Moles removed $=\frac{10^{21}}{6.02	imes10^{23}}=1.661	imes10^{-3}$ mol. Initial moles $=2.8	imes10^{-3}+1.661	imes10^{-3}=4.461	imes10^{-3}$ mol. Mass $=4.461	imes10^{-3}	imes44=196.3\approx196.2$ mg. Answer (3).

Q. $2.8\times10^{-3}$ mol of $\mathrm{CO_2}$ is left after removing $10^{21}$ molecules from its ' $x$ ' mg sample. The mass of $\mathrm{CO_2}$ taken initially is Given: $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$

Explanation

Reviewed by Dr. Abubacker

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