Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYEasy

Which of the following contains the maximum number of atoms?

1 g of Mg(s)

1 g of O₂ (g)

1 g of Li(s)

1 g of Ag(s)

Principle: The number of atoms ( $N$ ) in a given mass ( $w$ ) of a substance is calculated using the mole concept: $N = \text{Moles} \times N_A \times \text{Atomicity}$ $N = \frac{w}{\text{Molar Mass}} \times N_A \times \text{Atomicity}$ 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).
Calculations (using approximate atomic masses):

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

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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