How many Cl^- ions are present in 400.5 grams of AlCl_3?

• A. 7.23 × 10^{23} Cl^- ions
• B. 1.81 × 10^{24} Cl^- ions
• C. 2.42 × 10^{25} Cl^- ions
• D. 5.44 × 10^{24} Cl^- ions

Answer: (D)

Verified step by step guidance

1. Calculate the molar mass of aluminum chloride (AlCl_3) by adding the atomic masses of 1 aluminum (Al) atom and 3 chlorine (Cl) atoms: \(\text{Molar mass of AlCl}_3 = (1 \times M_{Al}) + (3 \times M_{Cl})\).
2. Determine the number of moles of AlCl_3 in 400.5 grams by dividing the given mass by the molar mass: \(\text{moles of AlCl}_3 = \frac{400.5}{\text{molar mass of AlCl}_3}\).
3. Recognize that each formula unit of AlCl_3 contains 3 chloride ions (Cl^-), so multiply the moles of AlCl_3 by 3 to find the moles of Cl^- ions: \(\text{moles of Cl}^- = 3 \times \text{moles of AlCl}_3\).
4. Use Avogadro's number (\$6.022 \times 10^{23}\( entities/mol) to convert moles of Cl^- ions to the number of Cl^- ions: \)\text{number of Cl}^- = \text{moles of Cl}^- \times 6.022 \times 10^{23}$.
5. Express the final answer in scientific notation to match the format of the given options.