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QUESTION 4 A 500.0 mL water sample is titrated with 15.00 mLof 0.1040 M EDTA. How many mg of Caco3 are in the sample?
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Answer #1

millimoles of EDTA = 15 x 0.1040 = 1.56

millimoles of EDTA = millimoles of CaCO3

moles of CaCO3 = 1.56 x 10^-3

mass of CaCO3 = 1.56 x 10^-3 x 100

                          = 0.156 g

mass of CaCO3 = 156 mg

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