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74.5513 g/mol KCl * * 1 mol/eq * 0.1 eq/L = 7.455 g/L

So accurately weight 7.455 g dry, analytical grade KCl and dissolve in volumetric flask of exact 1.000 Liter and fill up with water to the 1L-mark, close and carefully mix by repeated inversion. In this way you can get an exact 0.1000 N KCl standard solution.

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