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(A) To find the number of atoms,we calculate the moles and multiply by the atomicity and Avogadro's number $(N_A = 6.023 	imes 10^{23} \ mol^{-1})$.$

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$4.0 \ g$ of hydrogen

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(A) To find the number of atoms,we calculate the moles and multiply by the atomicity and Avogadro's number $(N_A = 6.023 	imes 10^{23} \ mol^{-1})$.$A$) $4.0 \ g$ of $H_2$: Moles $= \frac{4}{2} = 2 \ mol$. Atoms $= 2 	imes 2 	imes N_A = 4 N_A$.$B$) $71.0 \ g$ of $Cl_2$: Moles $= \frac{71}{71} = 1 \ mol$. Atoms $= 1 	imes 2 	imes N_A = 2 N_A$.$C$) $127.0 \ g$ of $I_2$: Moles $= \frac{127}{254} = 0.5 \ mol$. Atoms $= 0.5 	imes 2 	imes N_A = 1 N_A$.$D$) $48.0 \ g$ of $Mg$: Moles $= \frac{48}{24} = 2 \ mol$. Atoms $= 2 	imes 1 	imes N_A = 2 N_A$.Comparing the values,$4.0 \ g$ of hydrogen has the largest number of atoms $(4 N_A)$.

Number of atoms in the following samples of substances is largest in

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