Question

a container holds 1.0 l of 0.5 m alcl₃ solution. how many moles of cl⁻ ions are in the solution?
alcl₃ → al³⁺ + 3cl⁻
1.5 moles cl⁻
3.0 moles cl⁻
0.5 moles cl⁻

Explanation:

Step1: Calculate moles of \( \text{AlCl}_3 \)

Molarity formula: \( M = \frac{n}{V} \), so \( n = M \times V \).
Given \( M = 0.5 \, \text{M} \), \( V = 1.0 \, \text{L} \).
\( n_{\text{AlCl}_3} = 0.5 \, \text{mol/L} \times 1.0 \, \text{L} = 0.5 \, \text{mol} \).

Step2: Relate moles of \( \text{AlCl}_3 \) to \( \text{Cl}^- \)

From dissociation: \( \text{AlCl}_3
ightarrow \text{Al}^{3+} + 3\text{Cl}^- \).
1 mol \( \text{AlCl}_3 \) produces 3 mol \( \text{Cl}^- \).
So \( n_{\text{Cl}^-} = 3 \times n_{\text{AlCl}_3} = 3 \times 0.5 \, \text{mol} = 1.5 \, \text{mol} \).

Answer:

1.5 moles Cl⁻