Posted by Anonymous on Sunday, February 28, 2010 at 12:58am.
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
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