Exactly of a aqueous solution of is mixed with of an aqueous solution of unknown molarity. The resul — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
Exactly $40\text{ mL}$ of a $2.0\text{ M}$ aqueous solution of $CaCl_2$ is mixed with $60\text{ mL}$ of an aqueous $AlCl_3$ solution of unknown molarity. The resulting mixture is then diluted with pure water to a final total volume of $500\text{ mL}$. If this diluted solution contains exactly $0.70\text{ moles}$ of chloride ions ($Cl^-$), what was the exact initial molarity of the $AlCl_3$ solution used?
💡 Solution & Explanation
Step 1: Set up the total moles of $Cl^-$ equation. Total moles of $Cl^- = (\text{Moles } Cl^- \text{ from } CaCl_2) + (\text{Moles } Cl^- \text{ from } AlCl_3)$. Step 2: Calculate $Cl^-$ from $CaCl_2$. Moles = Volume(L) $\times$ Molarity $\times$ (Number of $Cl^-$ per molecule) = $0.040\text{ L} \times 2.0\text{ M} \times 2 = 0.16\text{ moles}$. Step 3: Calculate $Cl^-$ from $AlCl_3$ (let molarity be M). Moles = $0.060\text{ L} \times M \times 3 = 0.18 M\text{ moles}$. Step 4: Equate to total $Cl^-$ and solve. $0.16 + 0.18 M = 0.70 \implies 0.18 M = 0.54 \implies M = \frac{0.54}{0.18} = 3.0\text{ M}$.