0.2108g CaCO2 x (1 mol/molar mass CaCO3) = moles CaCO3.
You took 10 of 100 mL for the aliquot which is 1/10 of the sample so moles from above x 0.1 = moles titrated. If the ratio is 1:1 then that will be the moles EDTA.
I need to determine the moles of EDTA in a 1:1 stoichemetric titration of a 10.0 mL aliquot of .2108 g of CaCO3 dissolved in 100ml of h2o please help
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