Posted by Anonymous on .
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.

Chemistry 
DrBob222,
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. 
Chemistry 
D.Or.,
DrBob222, how are you sure this is the right answer. could you explain a little bit more?

Chemistry 
Anonymous,
Relating Density to a 1:1 stoichiometry equation