How many moles of chloride ions are present in 250 mL of a 0.40 M solution of aluminum chloride, AlCl3?
Show the answer and explanation
Correct answer
B. 0.30 mol
Principle or equation
Molarity (M) = moles of solute per liter of solution. The formula AlCl3 indicates 3 chloride ions per formula unit. Moles of Cl- = molarity × volume (L) × 3.
Why this answer is correct
First, convert volume to liters: 250 mL = 0.250 L. Moles of AlCl3 = 0.40 M × 0.250 L = 0.10 mol. Since each mole of AlCl3 gives 3 moles of Cl-, moles of Cl- = 0.10 mol × 3 = 0.30 mol.
Example
For 100 mL of 0.50 M CaCl2, moles of Cl- = 0.50 × 0.100 × 2 = 0.10 mol.
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.