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Chemistry
Posted 10 months ago
9) with (30+Y) gCa3(PO4)2\mathrm{g} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} (molar mass =310 g/mol)\left.=310 \mathrm{~g} / \mathrm{mol}\right), what is the maximum volume of the solution with a concentration of 0.375M0.375 \mathrm{M} ?
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Answer from Sia
Posted 10 months ago
Solution
1
Calculate the moles of $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$: Given the molar mass of Ca3(PO4)2\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 is 310 g/mol310 \mathrm{~g/mol}, calculate the moles of Ca3(PO4)2\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 using the mass provided (30+Y)g(30+Y) \mathrm{g}. The formula to calculate moles is moles=massmolar mass\text{moles} = \frac{\text{mass}}{\text{molar mass}}
2
Determine the volume of solution: Using the concentration 0.375M0.375 \mathrm{M}, which means 0.3750.375 moles of solute per liter of solution, calculate the maximum volume of the solution that can be prepared. The formula to calculate volume is volume=moles of soluteconcentration\text{volume} = \frac{\text{moles of solute}}{\text{concentration}}
Answer
The maximum volume of the solution with a concentration of 0.375M0.375 \mathrm{M} that can be prepared from (30+Y)g(30+Y) \mathrm{g} of Ca3(PO4)2\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 is (30+Y)/3100.375\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.

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