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JFernando
Answers (1)
first get for the moles of CaCO3 mols CaCO3= 0.5662g*(1mol CaCO3/100gCaCO3) Molarity of CaCO3= mols CaCO3/L of solution =0.005622mols/1L M EDTA= [(molarity CaCO3)(mL aliquot)]/ mL of EDTA M EDTA= [(0.005622)(25)]/21.88 = 0.00642 M
