Solve for the normality of 0.545 grams sodium carbonate in a 500 ml solution.

Answer 1

Na2CO3 ==> 2Na+ + CO32-

each mole of Na2CO3 has 2 equivalents

molar mass = 106 g/mol

0.545 g Na2CO3 x 1 mol / 106 g = 0.00514 mols

0.00514 mols x 2 equivalents / mole = 0.0103 equivalents

N = equivalents / liter = 0.01013 equivalents / 0.5 L

Normality = 0.0206 N