Which of the following contains the maximum number of atoms?

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Accepted Answer

1. **Principle:** The number of atoms ($N$) in a given mass ($w$) of a substance is calculated using the mole concept:
   $$ N = 	ext{Moles} 	imes N_A 	imes 	ext{Atomicity} $$
   $$ N = \frac{w}{	ext{Molar Mass}} 	imes N_A 	imes 	ext{Atomicity} $$
   Since the mass ($w$) is constant (1 g) for all options, the number of atoms is inversely proportional to the atomic mass (or effective mass per atom).

2. **Calculations (using approximate atomic masses):**
   - **Option A (Mg):** Atomic mass $\approx 24$ g/mol.
     $$ N_{Mg} = \frac{1}{24} 	imes 1 	imes N_A \approx 0.04 N_A $$
   - **Option B ($O_2$):** Molecular mass = 32 g/mol. Atomicity = 2.
     $$ N_{O} = \frac{1}{32} 	imes 2 	imes N_A = \frac{1}{16} N_A \approx 0.06 N_A $$
   - **Option C (Li):** Atomic mass $\approx 7$ g/mol.
     $$ N_{Li} = \frac{1}{7} 	imes 1 	imes N_A \approx 0.14 N_A $$
   - **Option D (Ag):** Atomic mass $\approx 108$ g/mol.
     $$ N_{Ag} = \frac{1}{108} 	imes 1 	imes N_A \approx 0.009 N_A $$

3. **Conclusion:** Lithium (Li) has the lowest atomic mass among the given monoatomic solids and the effective atomic mass of oxygen in $O_2$ (16). Therefore, 1 g of Li contains the maximum number of atoms .

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