posted by Anonymous .
A 15.00 ml sample of a standard solution containing 1 g of CaCO3/L required 8.45 ml of EDTA to fully complex the Ca present. Calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCO3.'
Sorry, forgot to write the entire question.
1 g CaCO3/L x 0.015 L = 0.015 g or 15 mg CaCO3 in the aliquot.
That is equivalent to 8.45 mL EDTA; therefore,
15 mg/8.45 mL = 1.775 mg CaCO3/mL EDTA. I wold round that to 1.78.
DrBob222, how are you sure this is the right answer. could you explain a little bit more?
Relating Density to a 1:1 stoichiometry equation