Question Detail

Moles Cl⁻ = (3.0 × 10²³)/(6.022 × 10²³) = 0.498 mol. Each AlCl₃ gives 3 Cl⁻, so moles AlCl₃ = 0.498/3 = 0.166 mol. Molar mass AlCl₃ = 27 + 106.5 = 133.5 g/mol. Mass = 0.166 × 133.5 = 22.2 g. Wait, that gives 22.2 g. The original pattern gave 78.9 g for MgCl₂ with 5×10²³ Mg²⁺. For 3×10²³ Cl⁻, the answer is 22.2 g. The option 44.4 g would be for 6×10²³ Cl⁻. Given options, 22.2 g is A.