Genecks Posted March 22, 2012 Share Posted March 22, 2012 (edited) Say I'm making a solution of 3 M KCl. I did that today. I think... Anyway, I made about 50 mL of KCl for a pH electrode solution. Now, if I remember correctly, (74.56 g/mol) / 1 L = 1 M solution of KCl 3 * "" with 1 L volume used = 3 M KCl But.. it's not that I'm adding 1 L of ddH2O, right? I put in the KCl, and I add enough water so that it appears as though 1 L of volume has been taken up by both the KCl and the water, right? So, when thinking about that 1 L of volume, I'm also considering the volume the KCl takes up. Edited March 22, 2012 by Genecks Link to comment Share on other sites More sharing options...
mississippichem Posted March 23, 2012 Share Posted March 23, 2012 (edited) Right. Molar concentration is defined as moles of solute per liter of solution. This should be distinguished from liters of solvent. A side note: Your equation is not dimensionally correct. [math] \frac{3 \ \mathrm{mol}}{1 \ \mathrm{L}}= 3 \ \mathrm{ M} [/math] Edited March 23, 2012 by mississippichem Link to comment Share on other sites More sharing options...
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