Making a molar solution: How does the volume aspect work?

[Genecks]

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

[mississippichem]

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