9) with (30+Y) $\mathrm{g} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ (molar mass $\left.=310 \mathrm{~g} / \mathrm{mol}\right)$, what is the maximum volume of the solution with a concentration of $0.375 \mathrm{M}$ ?

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∻Solution∻ ‖ 1 ⋮ **Calculate the moles of $\mathrm{Ca}_3\left(\mathrm{PO}_4 ight)_2$**: Given the molar mass of $\mathrm{Ca}_3\left(\mathrm{PO}_4 ight)_2$ is $310 \mathrm{~g/mol}$, calculate the moles of $\mathrm{Ca}_3\left(\mathrm{PO}_4 ight)_2$ using the mass provided $(30+Y) \mathrm{g}$. The formula to calculate moles is $ ext{moles} = \frac{ ext{mass}}{ ext{molar mass}}$. ‖ ‖ 2 ⋮ **Determine the volume of solution**: Using the concentration $0.375 \mathrm{M}$, which means $0.375$ moles of solute per liter of solution, calculate the maximum volume of the solution that can be prepared. The formula to calculate volume is $ ext{volume} = \frac{ ext{moles of solute}}{ ext{concentration}}$. ‖ ∻Answer∻ ⚹ The maximum volume of the solution with a concentration of $0.375 \mathrm{M}$ that can be prepared from $(30+Y) \mathrm{g}$ of $\mathrm{Ca}_3\left(\mathrm{PO}_4 ight)_2$ is $\frac{(30+Y) / 310}{0.375}$ liters. ⚹ ∻Key Concept∻ ⚹ Molarity and solution preparation ⚹ ∻Explanation∻ ⚹ Molarity (M) is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution. To find the maximum volume of a solution with a given molarity, one must first calculate the moles of solute available and then divide by the molarity. ⚹◊What is the formula to calculate the maximum volume of a solution with a given concentration?⍭ Generate me a similar question◊

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