Exactly 50.0ml solution of Na2CO3 was titrated with 65.8ml of 3.00M HCl. If the specific gravity of the Na2CO3 solution is 1.25, what percent by weight of Na2CO3 does it contain?

Answer 1

I assume you titrated ALL of the Na2CO3; i.e., Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols HCl = M x L = 3.00 x 0.0658 = approximately 0.2 (you can do it more accurately).
mols Na2CO3 = 1/2 that (from the coefficients) = approximately 0.1
g Na2CO3 = mols x molar mass = 0.1 x 106 = approximately 10 g.
mass of the 50 mL soln = 50 x 1.25 = 62.5 grams.
% Na2CO3 = (10/62.5)*10 = ?%
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