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

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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 by Genecks

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 by mississippichem

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