How many chloride ions are present in 65.5 mL of 0.210 M AlCl_3 solution?

• A. 7.98 × 10^{21} chloride ions
• B. 1.24 × 10^{22} chloride ions
• C. 2.48 × 10^{22} chloride ions
• D. 3.94 × 10^{22} chloride ions

Answer: (C)

Verified step by step guidance

1. Identify the given information: volume of AlCl_3 solution = 65.5 mL, molarity (M) of AlCl_3 = 0.210 M. Remember to convert volume from milliliters to liters by dividing by 1000.
2. Calculate the number of moles of AlCl_3 in the given volume using the formula: \(\text{moles of AlCl}_3 = M \times V\), where \(V\) is in liters.
3. Determine the number of moles of chloride ions. Since each formula unit of AlCl_3 contains 3 chloride ions, multiply the moles of AlCl_3 by 3 to get moles of Cl\(^-\) ions.
4. Use Avogadro's number, \$6.022 \times 10^{23}\( particles/mole, to convert moles of chloride ions to the number of chloride ions: \)\text{number of Cl}^- = \text{moles of Cl}^- \times 6.022 \times 10^{23}$.
5. Perform the calculations stepwise to find the total number of chloride ions present in the 65.5 mL of 0.210 M AlCl_3 solution.