Which of the following has the maximum number | vedclass

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(D) To find the number of atoms,we first calculate the number of moles $(n = 	ext{mass} / 	ext{molar mass})$ and then multiply by the number of atom

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$18 \, g$ of $CH_4$

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(D) To find the number of atoms,we first calculate the number of moles $(n = 	ext{mass} / 	ext{molar mass})$ and then multiply by the number of atoms per molecule and Avogadro's number $(N_A)$.$A) \, 18 \, g$ of $H_2O$: Molar mass = $18 \, g/mol$. Moles = $18/18 = 1 \, mol$. Atoms per molecule = $3$ $(2H + 1O)$. Total atoms = $1 	imes 3 	imes N_A = 3N_A$.$B) \, 18 \, g$ of $O_2$: Molar mass = $32 \, g/mol$. Moles = $18/32 = 0.5625 \, mol$. Atoms per molecule = $2$. Total atoms = $0.5625 	imes 2 	imes N_A = 1.125N_A$.$C) \, 18 \, g$ of $CO_2$: Molar mass = $44 \, g/mol$. Moles = $18/44 \approx 0.409 \, mol$. Atoms per molecule = $3$ $(1C + 2O)$. Total atoms = $0.409 	imes 3 	imes N_A \approx 1.227N_A$.$D) \, 18 \, g$ of $CH_4$: Molar mass = $16 \, g/mol$. Moles = $18/16 = 1.125 \, mol$. Atoms per molecule = $5$ $(1C + 4H)$. Total atoms = $1.125 	imes 5 	imes N_A = 5.625N_A$.Comparing the values,$CH_4$ has the maximum number of atoms.

Which of the following has the maximum number of atoms?

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