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A solution is prepared by dissolving 28 g of Ca3(PO4)2 (MW = 310.18 g/mol) in enough water to make a 150 mL of solution. What is the molarity of Ca2+ ions in solution? (in M)

Question

A solution is prepared by dissolving 28 g of Ca3(PO4)2 (MW = 310.18 g/mol) in enough water to make a 150 mL of solution. What is the molarity of Ca2+ ions in solution? (in M)

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Solution

Step 1: Calculate the number of moles of Ca3(PO4)2 The number of moles = mass (g) / molar mass (g/mol) = 28 g / 310.18 g/mol = 0.0902 mol

Step 2: Calculate the volume of the solution in liters 150 mL = 150 / 1000 = 0.15 L

Step 3: Calculate the molarity of Ca3(PO4)2 Molarity = moles / volume (L) = 0.0902 mol / 0.15 L = 0.601 M

Step 4: Calculate the molarity of Ca2+ ions Each formula unit of Ca3(PO4)2 produces 3 Ca2+ ions. Therefore, the molarity of Ca2+ ions is 3 times the molarity of Ca3(PO4)2. = 3 * 0.601 M = 1.803 M

So, the molarity of Ca2+ ions in the solution is 1.803 M.

This problem has been solved

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