Total simple ions from dissociation of CuSO₄ formula units Steps:

• Molar mass of anhydrous CuSO₄ is 160 g/mol (Cu 64 + S 32 + 4×O 16), so 16.0 g = 16/160 = 0.1 mol.
• Number of CuSO₄ units = 0.1 mol × 6.02 × 10²³ = 6.02 × 10²².
• Each CuSO₄ provides 6 simple ions: 1 Cu²⁺, 1 S⁶⁺, 4 O²⁻.
• Total ions = 6 × 6.02 × 10²² = 3.61 × 10²³ (matches A, using 6.0 × 10²³ for Avogadro's number approximation).

Why A is correct:

• A uses the formula for total simple ions (6 per CuSO₄) times moles times Avogadro's number, reflecting complete dissociation into atomic ions.

Why the others are wrong:

• B overestimates by factor of 10, likely from using 1 mol instead of 0.1 mol.
• C undercounts ions, treating CuSO₄ as only 2 ions (Cu²⁺ + SO₄²⁻) without breaking down SO₄²⁻.
• D overestimates by factor of 10, likely from 6 ions but using 1 mol.

Final answer: A