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Conversion Factors  Molarity
Dr. MJ Patterson
The latest conversion factor to pop up is molarity, or the concentration of a solution in moles of solute per liter of solution. We can write a conversion factor from the molarity because the numerical value tells us how many moles of solute are in exactly 1 liter of solution. For example, a 0.1 M HCl_{(}_{aq}_{)} solution means that in 1 liter, there are 0.1 moles of HCl_{(aq)}. The conversion factor is:
1 liter solution = 0.1 moles HCl_{(}_{aq}_{)}
We now have a way to convert between moles of solute and the volume of the solution. This is convenient because it is far easier to meaure the volume of a solution in a graduated cylinder than it is to weigh a solution. Before, we could only convert from mass to moles using the molar mass. This new conversion factor will let us convert from a second experimentally measurable quantity  the volume.
Notation note: You will frequently see a compound name or formula written in square brackets, such as [HCl] = 1.1 M. The square brackets indicate the concentration in molarity of whatever is inside the brackets. This example would be read as "The concentration of hydrochloric acid is 1.1 M."
Example 1:
How many moles of salt are present in 0.500 L of 1.0 M saline (aqueous salt solution)? How many grams of salt?
Solution 1:
We will use the molarity as a conversion factor, so we must start with the 0.500 L of solution:
1 L saline = 1.0 mol NaCl
(0.500 L saline)  x  (1.0 mol NaCl)  = 0.50 mol NaCl 
(1)  (1 L saline) 


To determine how many grams of salt are in the sample, we need to use the molar mass (58.44 g/mol) to convert from moles of NaCl to grams.
(0.50 mol NaCl)  x  (58.44 g NaCl)  = 29.22 g NaCl = 29 g NaCl 
(1)  (1 mol NaCl) 


Ion Concentration:
Ionic compounds dissolved in water can easily give ion concentrations that are higher than the stated concentration for the compound. That is because the convention that is followed in chemistry is to give the concentration of the compound as if it were dissolved as complete molecules. If it breaks down into ions upon dissolving in water, the ion concentration could be different from the stated concentration, multiplied by the stoichiometric ratio.
Example 2:
What is the concentation of each ion in 1.2 M Ca(NO_{3})_{2(aq)}? What is the total concentration of ions?
Solution 2:
When calcium nitrate dissolves in water, each formula unit releases a single calcium ion and two nitrate ions:
Ca(NO_{3})_{2(aq) } => Ca^{2+}_{(aq)} + 2NO_{3}^{}_{(aq)}
If the solution is 1.2 M in Ca(NO_{3})_{2}, then it will also be 1.2 M in Ca^{2+}. But, since there are twice as many nitrate ions, it will be 2(1.2) = 2.4 M in NO_{3}^{}.
To find the total ion concentration, note that one formula unit of Ca(NO_{3})_{2} will produce 3 ions  one cation and two anions. Therefore, the total ion concentration is 3(1.2) = 3.6 M.