Dissolved 0.7500 g calcium chloride dihydrate and impure, in distilled water and the volume was elevated to 200.00 mL in a volumetric flask. A rate of 10.00 mL of the same consumed the titration 18.9 mL 0.0125 M EDTA using CALCON as indicator. What is the purity of the salt (% w/w)

Answer 1

1 mole EDTA reacts with 1 mol CaCl2.2H2O.
mols EDTA = M x L = 0.0125 x 0.0189 = approx 0.00024 but you need a better answer to that as well all of the steps that follow. They are estimates also.
This was in 10 mL aliquot from 200 mL total; therefore 0.00024 x (200/10) = approx 0.0045 mols CaCl2.2H2O in 0.75 g sample. Convert mols CaCl2.2H2O to grams CaCl2.2H2O, then
% purity = (g CaCl2.2H2O/g sample)*100 = ?