1. Calculate moles of AlCl3:

* Molarity (M) = moles of solute / liters of solution

* Liters of solution: Convert 65.5 mL to liters by dividing by 1000: 65.5 mL / 1000 mL/L = 0.0655 L

* Moles of AlCl3: (0.210 M) * (0.0655 L) = 0.013755 moles AlCl3

2. Determine moles of chloride ions:

* AlCl3 dissociates into 1 Al³⁺ ion and 3 Cl⁻ ions.

* Moles of Cl⁻: (0.013755 moles AlCl3) * (3 moles Cl⁻ / 1 mole AlCl3) = 0.041265 moles Cl⁻

3. Calculate the number of chloride ions:

* Avogadro's number: 6.022 x 10²³ ions/mole

* Number of Cl⁻ ions: (0.041265 moles Cl⁻) * (6.022 x 10²³ ions/mole) = 2.48 x 10²² chloride ions

Therefore, there are approximately 2.48 x 10²² chloride ions present in 65.5 mL of 0.210 M AlCl3 solution.