Question

A 47.30 mL aliquot from a 0.485 L solution that contains 0.465 g of MnSO4 (MW=151.00 g/mol) required 39.4 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO3 ( MW=100.09 g/mol) will react with 1.78 mL of the EDTA solution?

A 47.30 mL aliquot from a 0.485 L solution that contains 0.465 g of MnSO4 (MW = 151.00 g/mol) required 39.4 mL of an EDTA sol

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a moles. 0.465 gom Mnog ie, 0.465 m 151 23.08 × 10-3 moles 3. 47.30 ml allaust contains : 3.08*10-3 x 0.0473 ml 0.48 5L 23x10

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