10 g each of CO_2,NH_3,and O_2 were taken in | vedclass

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(D) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{moles} 	imes N_A 	imes 	ext{atomicity}$.$1$. For $CO_2$: $	ext{moles} =

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$NH_3 > CO_2 > O_2$

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(D) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{moles} 	imes N_A 	imes 	ext{atomicity}$.$1$. For $CO_2$: $	ext{moles} = \frac{10}{44} \approx 0.227$. $	ext{Atomicity} = 3$. $	ext{Atoms} = 0.227 	imes 3 	imes N_A = 0.681 N_A$.$2$. For $NH_3$: $	ext{moles} = \frac{10}{17} \approx 0.588$. $	ext{Atomicity} = 4$. $	ext{Atoms} = 0.588 	imes 4 	imes N_A = 2.352 N_A$.$3$. For $O_2$: $	ext{moles} = \frac{10}{32} = 0.3125$. $	ext{Atomicity} = 2$. $	ext{Atoms} = 0.3125 	imes 2 	imes N_A = 0.625 N_A$.Comparing the values: $2.352 N_A (NH_3) > 0.681 N_A (CO_2) > 0.625 N_A (O_2)$.Thus,the decreasing order is $NH_3 > CO_2 > O_2$.

$10 \ g$ each of $CO_2$,$NH_3$,and $O_2$ were taken in three separate flasks. What is the correct decreasing order of the number of atoms?

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