What precise volume of pure water must be added to a mixture of of and of to obtain a resulting solu — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Total millimoles $= M_1 V_1 + M_2 V_2 = (250 \times 6) + (650 \times 3) = 1500 + 1950 = 3450\text{ mmol}$. For a $3\text{ M}$ solution, required final volume $V = 3450 / 3 = 1150\text{ mL}$. Initial volume $= 250 + 650 = 900\text{ mL}$. Volume of water to add $= 1150 - 900 = 250\text{ mL}$.