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(C) $1$. Calculate the number of atoms in $6 \,g$ of $H_2O$: $n(H_2O) = \frac{6 \,g}{18 \,g/mol} = \frac{1}{3} \,mol$. Since each $H_2O$ molecule has

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$1 \,g$ $H_2$

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(C) $1$. Calculate the number of atoms in $6 \,g$ of $H_2O$: $n(H_2O) = \frac{6 \,g}{18 \,g/mol} = \frac{1}{3} \,mol$. Since each $H_2O$ molecule has $3$ atoms ($2$ $H$ and $1$ $O$), total atoms $= 3 	imes \frac{1}{3} \,N_A = N_A$. $2$. Evaluate options: $A$: $n(He) = \frac{0.4}{4} = 0.1 \,mol$. Atoms $= 0.1 \,N_A$. $B$: $n(CO_2) = \frac{22}{44} = 0.5 \,mol$. Atoms $= 3 	imes 0.5 \,N_A = 1.5 \,N_A$. $C$: $n(H_2) = \frac{1}{2} = 0.5 \,mol$. Atoms $= 2 	imes 0.5 \,N_A = N_A$. $D$: $n(CO) = \frac{12}{28} \approx 0.43 \,mol$. Atoms $= 2 	imes 0.43 \,N_A = 0.86 \,N_A$. Thus, $1 \,g$ of $H_2$ has the same number of atoms.

Which one of the following has the same number of atoms as are in $6 \,g$ of $H_2O$?

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