To determine which substance has the maximum number of atoms, we need to calculate the number of moles for each substance and then find the number of atoms.$\\$The number of moles $n$ is given by:$\\ n = \frac{mass}{molar \ mass} \\$Let's calculate for each option:$\\$Option 1: 1 g of Mg(s)$\\$Molar mass of Mg$= 24 \,$g/mol$\\ n_{Mg} = \frac{1}{24} \,$mol$\\$Number of atoms in Mg$= n_{Mg} \times N_A = \frac{1}{24} \times N_A \\$Option 2: 1 g of $O_2(g) \\$Molar mass of $O_2 = 32 \,$g/mol$\\ n_{O_2} = \frac{1}{32} \,$mol$\\$Number of molecules in $O_2 = n_{O_2} \times N_A = \frac{1}{32} \times N_A \\$Number of atoms in $O_2 = 2 \times \frac{1}{32} \times N_A = \frac{1}{16} \times N_A \\$Option 3: 1 g of Li(s)$\\$Molar mass of Li$= 7 \,$g/mol$\newline n_{Li} = \frac{1}{7} \,$mol$\\$Number of atoms in Li$= n_{Li} \times N_A = \frac{1}{7} \times N_A \\$Option 4: 1 g of Ag(s)$\\$Molar mass of Ag$= 108 \,$g/mol$\\ n_{Ag} = \frac{1}{108} \,$mol$\\$Number of atoms in Ag$= n_{Ag} \times N_A = \frac{1}{108} \times N_A \\$Comparing the number of atoms:$\\$• Mg: $\frac{1}{24} \times N_A \\$• $O_2: \frac{1}{16} \times N_A \\$• Li: $\frac{1}{7} \times N_A \\$• Ag: $\frac{1}{108} \times N_A \\$The maximum number of atoms is in Li, as $\frac{1}{7} > \frac{1}{24}, \frac{1}{16}, \frac{1}{108}. \\$Therefore, the correct option is Option 3: 1 g of Li(s).