From the following,the largest number of atom | vedclass

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(C) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{atomicity} 	imes 	ext{moles} 	imes N_A$.$A$. For $36 \ g \ H_2O$: $	ext{

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$46 \ g \ C_2H_5OH$

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(C) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{atomicity} 	imes 	ext{moles} 	imes N_A$.$A$. For $36 \ g \ H_2O$: $	ext{moles} = \frac{36}{18} = 2 \ mol$. $	ext{Atomicity} = 3$. $	ext{Total atoms} = 2 	imes 3 	imes N_A = 6 N_A$.$B$. For $28 \ g \ CO$: $	ext{moles} = \frac{28}{28} = 1 \ mol$. $	ext{Atomicity} = 2$. $	ext{Total atoms} = 1 	imes 2 	imes N_A = 2 N_A$.$C$. For $46 \ g \ C_2H_5OH$: $	ext{moles} = \frac{46}{46} = 1 \ mol$. $	ext{Atomicity} = 9$. $	ext{Total atoms} = 1 	imes 9 	imes N_A = 9 N_A$.$D$. For $54 \ g \ N_2O_5$: $	ext{moles} = \frac{54}{108} = 0.5 \ mol$. $	ext{Atomicity} = 7$. $	ext{Total atoms} = 0.5 	imes 7 	imes N_A = 3.5 N_A$.Comparing the values,$46 \ g \ C_2H_5OH$ contains the largest number of atoms.

From the following,the largest number of atoms are in

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