The answer is: 15.66 moles of chloride ions.
Dissociation of aluminium chloride in water: AlCl₃(aq) → Al³⁺(aq) + 3Cl⁻(aq).
n(AlCl₃) = 5.22 mol; amount of aluminium chloride.
From chemical reaction: n(AlCl₃) : n(Cl⁻) = 1 : 3.
n(Cl⁻) = 3 · 5.22 mol.
n(Cl⁻) = 15.66 mol; amount of chloride ions.
N(Cl⁻) = n(Cl⁻) · Na.
N(Cl⁻) = 15.66 mol · 6.022·10²³ 1/mol.
N(Cl⁻) = 9.43·10²⁴; number of chloride ions.
