How many chloride ions are present in 1.50 mol of aluminum chloride (AlCl_3)?

• A. 4.52 × 10^{24} chloride ions
• B. 2.71 × 10^{24} chloride ions
• C. 9.03 × 10^{23} chloride ions
• D. 1.50 × 10^{24} chloride ions

Answer: (A)

Verified step by step guidance

1. Identify the chemical formula of aluminum chloride, which is AlCl\_3. This means each formula unit contains 1 aluminum ion (Al^{3+}) and 3 chloride ions (Cl^{-}).
2. Determine the number of moles of chloride ions in 1.50 moles of AlCl\_3 by using the mole ratio from the formula. Since there are 3 chloride ions per formula unit, multiply the moles of AlCl\_3 by 3: \(1.50 \text{ mol AlCl\_3} \times 3 = 4.50 \text{ mol Cl}^-\).
3. Recall Avogadro's number, which is \(6.022 \times 10^{23}\) particles per mole. This number allows you to convert moles of ions to the actual number of ions.
4. Calculate the total number of chloride ions by multiplying the moles of chloride ions by Avogadro's number: \(4.50 \text{ mol Cl}^- \times 6.022 \times 10^{23} \text{ ions/mol} = \text{number of chloride ions}\).
5. Express the final answer in scientific notation, ensuring the units are chloride ions, to match the format of the provided options.