What is the concentration of chloride ions in a 2.50 m solution of aluminum chloride (AlCl_3)?

• A. 7.50 m
• B. 5.00 m
• C. 1.25 m
• D. 2.50 m

Answer: (A)

Verified step by step guidance

1. Identify the formula of aluminum chloride, which is AlCl\_3, indicating that each formula unit contains 1 aluminum ion (Al\^{3+}) and 3 chloride ions (Cl\^{-}).
2. Understand that the concentration given (2.50 m) refers to the molality of AlCl\_3, meaning 2.50 moles of AlCl\_3 per kilogram of solvent.
3. Since each mole of AlCl\_3 produces 3 moles of Cl\^{-} ions upon dissociation, multiply the molality of AlCl\_3 by 3 to find the molality of chloride ions.
4. Write the expression for chloride ion concentration as: \(\text{molality of Cl}^{-} = 3 \times 2.50\,m\).
5. Calculate the product to find the molality of chloride ions, which gives the concentration of Cl\^{-} in the solution.