Answer:
A) 1.68 x 10²⁸ Cu atoms.
Step-by-step explanation:
∵ The mass of Cu in the chalcopyrite ore = mass of the ore x weight fraction of Cu.
- Weight fraction of Cu = Weight % / 100 = 34.63 /100 = 0.3463.
∴ The mass of Cu in the chalcopyrite ore = mass of the ore x weight fraction of Cu = (5.11 x 10³ kg)(0.3463) = 1.77 x 10³ kg.
- We need to calculate the no. of moles of Cu atoms:
n of Cu = mass/atomic mass of Cu = (1770 x 10³ g) / (63.54 g/mol) = 27.853 x 10³ mol.
- We have that every mole of an element contains Avogadro's no. of atoms (6.022 x 10²³).
∴ 27.853 x 10³ mol of Cu contains = (27.853 x 10³ mol)(6.022 x 10²³) = 1.677 x 10²⁸ Cu atoms ≅ 1.68 x 10²⁸ Cu atoms.