Which solution contains the largest number of chloride ions, $C^l-$, assuming complete dissociation and the same final volume for each solution?

• A. $100.0\mathrm{mL}$ of $0.40M$ $MgC{l}_2$
• B. $100.0\mathrm{mL}$ of $0.30M$ $CaC{l}_2$
• C. $100.0\mathrm{mL}$ of $0.50M$ $NaCl$
• D. $100.0\mathrm{mL}$ of $0.20M$ $AlC{l}_3$

Answer: (B)

Verified step by step guidance

1. Identify the number of chloride ions released per formula unit of each compound upon complete dissociation. For example, MgCl\_2 dissociates into one Mg^{2+} ion and two Cl^{-} ions, so it releases 2 chloride ions per formula unit.
2. Write the dissociation equations for each compound to confirm the number of chloride ions produced: - MgCl\_2 \(\rightarrow\) Mg^{2+} + 2Cl^{-} - CaCl\_2 \(\rightarrow\) Ca^{2+} + 2Cl^{-} - NaCl \(\rightarrow\) Na^{+} + Cl^{-} - AlCl\_3 \(\rightarrow\) Al^{3+} + 3Cl^{-}
3. Calculate the total moles of each compound in 100.0 mL (0.100 L) of solution using the molarity: \(moles = Molarity \times Volume = M \times 0.100\) L
4. Multiply the moles of each compound by the number of chloride ions released per formula unit to find the total moles of Cl^{-} ions: \(total\ Cl^{-} = moles\ of\ compound \times number\ of\ Cl^{-} \ ions\ per\ formula\ unit\)
5. Compare the total moles of chloride ions for each solution to determine which contains the largest number of chloride ions.