Mn+2+ EDTA ---> MnY-2
Molar ratio of Mn+2 (MnSO4) and EDTA = 1:1
Molarity of stock MnSO4 = No. of moles/Vol in lts = (0.525 g/151 g/mol)/0.505 l = 0.00689 M
No. of moles of aliquot = 0.00689 M x 48.80 lts/1000 = 0.336/1000 = 0.000336 moles
A molar ratio of Mn+2 (MnSO4) and EDTA = 1:1
Moles of EDTA = 0.000336 moles = 0.336 mmoles
Volume of EDTA = 40.8 ml
molarity of EDTA = No. of moles/Vol = 0.336 mmoles/40.8 ml = 0.00824 M
No. of moles of EDTA in 1.80 ml = 0.00824 M x 1.80 ml = 0.0148 mmoles
Ca+2 + EDTA ---> CaY-2
Molar ratio of Ca+2 (CaCO3) and EDTA = 1:1
Hence moles of Ca+2 (CaCO3) = 0.0148 mmoles
Molar mass of CaCO3 = 100.09 g/mol
Mass of CaCO3 = Moles x molar mass = 0.0148 mmoles x 100.09 g/mol = 1.48 mg
A 48.80 mL aliquot from a 0.505 L solution that contains 0.525 g of MnSOg (MW...
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explanation A 47.90 mL aliquot from a 0.510 L solution that contains 0.485 g of MnSO4 (Fw 151.00 g/mol) required 35.0 mL of an EDTA solution to reach the end point in a titration. What mass (in milligrams) of CaCO3 (FW 100.09 g/mol) will react with 1.63 m of the EDTA solution? Number Caco, mass 1406.18 mg
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