of a aqueous solution of Magnesium chloride () is mixed with of a aqueous solution of Aluminium chlo — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
$15\text{ mL}$ of a $0.20\text{ M}$ aqueous solution of Magnesium chloride ($MgCl_2$) is mixed with $45\text{ mL}$ of a $0.40\text{ M}$ aqueous solution of Aluminium chloride ($AlCl_3$). Assuming complete dissociation of both salts, what is the final molarity of chloride ions ($Cl^-$) in the resulting mixture?
💡 Solution & Explanation
Step 1: Calculate the millimoles of $Cl^-$ from $MgCl_2$. $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$. Millimoles of $Cl^- = 15\text{ mL} \times 0.20\text{ M} \times 2 = 6\text{ mmol}$. Step 2: Calculate the millimoles of $Cl^-$ from $AlCl_3$. $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$. Millimoles of $Cl^- = 45\text{ mL} \times 0.40\text{ M} \times 3 = 54\text{ mmol}$. Step 3: Calculate the total millimoles and final volume. Total $Cl^- = 6 + 54 = 60\text{ mmol}$. Total Volume = $15 + 45 = 60\text{ mL}$. Step 4: Calculate final molarity. $[Cl^-] = \frac{60\text{ mmol}}{60\text{ mL}} = 1.0\text{ M}$.