1. **State the problem:** We want to find how many moles of Al₂(SO₄)₃ are present in 50.0 mL of 0.250 M Al₂(SO₄)₃ solution. 2. **Formula used:** Molarity (M) is defined as moles of solute per liter of solution: $$M = \frac{\text{moles of solute}}{\text{liters of solution}}$$ 3. **Rearrange the formula to find moles:** $$\text{moles of solute} = M \times \text{liters of solution}$$ 4. **Convert volume from mL to L:** $$50.0\ \text{mL} = \frac{50.0}{1000} = 0.0500\ \text{L}$$ 5. **Calculate moles of Al₂(SO₄)₃:** $$\text{moles} = 0.250\ \text{M} \times 0.0500\ \text{L} = 0.0125\ \text{moles}$$ **Final answer:** There are **0.0125 moles** of Al₂(SO₄)₃ in 50.0 mL of 0.250 M solution.