Which has the maximum number of molecules amo | vedclass

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(C) The number of molecules is calculated as: $	ext{Number of molecules} = 	ext{Moles} 	imes N_A$.For $A$: $	ext{Moles of } CO_2 = \frac{44 \ g}{4

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$8 \ g \ H_2$

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(C) The number of molecules is calculated as: $	ext{Number of molecules} = 	ext{Moles} 	imes N_A$.For $A$: $	ext{Moles of } CO_2 = \frac{44 \ g}{44 \ g/mol} = 1 \ mol$. Number of molecules = $1 	imes N_A$.For $B$: $	ext{Moles of } O_3 = \frac{48 \ g}{48 \ g/mol} = 1 \ mol$. Number of molecules = $1 	imes N_A$.For $C$: $	ext{Moles of } H_2 = \frac{8 \ g}{2 \ g/mol} = 4 \ mol$. Number of molecules = $4 	imes N_A$.For $D$: $	ext{Moles of } SO_2 = \frac{64 \ g}{64 \ g/mol} = 1 \ mol$. Number of molecules = $1 	imes N_A$.Comparing the values,$8 \ g \ H_2$ contains the maximum number of molecules $(4 	imes N_A)$.

Which has the maximum number of molecules among the following?

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