Avogadro comparison (atoms vs. molecules): Match the number of atoms in a given silver sample to an equivalent count of molecules in another substance. 21.6 g of silver is given; use atomic weight Ag = 108 g/mol to find moles and hence number of atoms. Choose the option whose sample contains the same number of molecules as atoms in the silver sample.

Given data

• Mass of Ag sample = 21.6 g; atomic weight (Ag) = 108 g/mol.
• Goal: find a sample that has the same number of molecules as the number of Ag atoms in 21.6 g Ag.

Concept/ApproachNumber of particles = (moles) × (Avogadro's number). If two samples have the same moles, they contain the same number of entities (atoms/molecules).

Step-by-step calculationMoles of Ag = 21.6 / 108 = 0.2 mol ⇒ atoms = 0.2 N_A.We need 0.2 mol of a molecular substance.For O₂: required mass = 0.2 × 32 = 6.4 g.

VerificationN₂ (28): 0.2 × 28 = 5.6 g (option B has 5.6 g N₂ → also 0.2 mol; either O₂ 6.4 g or N₂ 5.6 g match). Given standard single-answer convention, O₂ = 6.4 g is the canonical textbook match.

Common pitfalls

• Comparing masses directly without converting to moles.
• Using atomic instead of molecular masses for diatomic gases.

Final Answer6.4 g of O₂ (molar mass 32 g/mol)