Final answer:
A 32.4 g piece of sterling silver jewelry contains approximately 1.67 × 1022 silver atoms, after accounting for sterling silver's composition and converting grams to moles, then moles to atoms using Avogadro's number.
Step-by-step explanation:
To determine how many Ag atoms are present in a piece of sterling silver jewelry weighing 32.4 g, we must use the atomic mass of silver (Ag) and Avogadro's number. The atomic mass of silver is approximately 107.9 amu (atomic mass units), which means that one mole of silver has a mass of approximately 107.9 grams. Sterling silver is an alloy, typically made up of 92.5% silver and 7.5% other metals, usually copper. Therefore, first, we calculate the mass of pure silver in the piece:
Mass of pure Ag = 32.4 g × 92.5% = 29.97 g
Next, we use the molar mass of Ag to find the number of moles of Ag:
Moles of Ag = 29.97 g ÷ 107.9 g/mol ≈ 0.278 moles
Now we can use Avogadro's number, which is approximately 6.022 × 1023 atoms per mole, to calculate the number of Ag atoms:
Number of Ag atoms = 0.278 moles × 6.022 × 1023 atoms/mole ≈ 1.67 × 1022 atoms
So, a 32.4 g piece of sterling silver jewelry contains approximately 1.67 × 1022 silver atoms.