The maximum number of molecules is present in

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(C) To find the maximum number of molecules,we calculate the number of moles $(n)$ for each option:$A$: $n = \frac{0.5 \ g}{2 \ g/mol} = 0.25 \ mol$.

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$15 \ L$ of $H_2$ gas at $STP$

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(C) To find the maximum number of molecules,we calculate the number of moles $(n)$ for each option:$A$: $n = \frac{0.5 \ g}{2 \ g/mol} = 0.25 \ mol$. Molecules $= 0.25 	imes N_A = 0.25 N_A$.$B$: $n = \frac{10 \ g}{32 \ g/mol} = 0.3125 \ mol$. Molecules $= 0.3125 	imes N_A = 0.3125 N_A$.$C$: $n = \frac{15 \ L}{22.4 \ L/mol} \approx 0.67 \ mol$. Molecules $= 0.67 	imes N_A = 0.67 N_A$.$D$: $n = \frac{5 \ L}{22.4 \ L/mol} \approx 0.22 \ mol$. Molecules $= 0.22 	imes N_A = 0.22 N_A$.Comparing the values,$15 \ L$ of $H_2$ gas at $STP$ contains the maximum number of molecules.

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