What is the molarity of a solution of Ca3(PO4)2 that has a density of 1.25 g/mL and is 24.0% Ca3(PO4)2

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What is the molarity of a solution of Ca3(PO4)2 that has a density of 1.25 g/mL and is 24.0% Ca3(PO4)2

[From: ] [author: ] [Date: 12-12-15] [Hit: ]

M = 0.M= 0.......

Please show work, thank you!

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Assuming 1 L solution you have 240 mL Ca3(PO4)2 x 1.25 g/ml = 300 g Ca3(PO4)2

Molarity = moles / Liters

Molarity = grams of Ca3(PO4)2 divided by molar mass of Ca3(PO4)2 divided by 1 Liter
M = 300 g Ca3(PO4)2 / 310.18 g/mol Ca3(PO4)2
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1 L

M = 0.967180347 / 1 L
M = 0.967180347
M= 0.967 rounded to 3 SF

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