MicrobiologyBytes: Maths & Computers for Biologists: Molarities & Dilutions  Updated: October 19, 2004  Search 
Further information on this topic can be found in Chapter 4 of:
Maths from Scratch for Biologists
Numerical ability is an essential skill for everyone
studying the biological sciences but many students are frightened by
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mathematical skills in their chosen field of study. Maths from Scratch
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The mole (from "molecule") is the
SI (Système International) unit used to measure the amount of
a substance.
(symbol = mol)
It is also a small furry animal.
Technically, one mole is:
Complicated, isn't it?
Fortunately, this was simplified for us by
Amedeo Avogadro, who in 1811 the first person to was clearly
make the distinction between the molecule and the atom.
Avogadro suggested that:
which is now known as Avogadro's Principle.
It was long after Avogadro that the idea of a mole was introduced. Since a molecular weight in grams (mole) of any substance contains the same number of molecules, then according to Avogadro's principle, the molar volumes of all gases should be the same. The number of molecules in one mole is now called Avogadro's number. Avogadro had no knowledge of moles, or of the number that was to bear his name, which was never actually determined by Avogadro himself.
Avogadro's number is:
How big is Avogadro's number?
Big, huh? 
A solution is a homogeneous mixture where all particles exist as individual molecules or ions.
The molarity of a solution is calculated by taking the moles of solute and dividing by the litres of solution:
Learn this equation! 
(Just to confuse you, the molality (m) (yes, that is spelt right) of a solution is the concentration measured as moles of solute per kilogram of solvent. For example, a 1 m (not a 1 M) NaCl solution contains 1 mole of NaCl per kilogram of water. Molalities are preferred over molarities in experiments that involve temperature changes of solutions, e. g. calorimetry and freezing point depression experiments. Molarity = moles per litre, molality = moles per kilogram. OK, we won't talk about molality any more.)
Example 1:What is the molarity of 2 moles of solute dissolved in 1 litre of solvent?
2 mol  
Molarity = 

= 2 mol L^{1}  = 2 M 
1 litre 
i.e. 2 moles/litre, or "2 molar".
Example 2:
What is the molarity of 0.75 moles of solute dissolved in 2.5 litres of solvent?
0.75 mol  
Molarity = 

= 0.3 M 
2.5 L 
Example 3:
What is the molarity of 40 grams of NaOH dissolved in 2 litres of solvent?
This calculation must be performed in two stages:
40 grams  

= 1 mol 
40 grams / mol 
b)
1 mol  

= 0.5 M 
2 L 
Learn these equations!

Example 4:
When 2 grams of NaCl (molecular weight 58.44 g mol^{1}) is dissolved in 100 mL of solute, what is the molarity of the solution? (MV=g/mol.wt.)
2 g  
( x M) (0.1 L) = 

58.44  
Therefore:  
(x M) (0.1) = 0.034  
= 0.34 M 
Example 5:
How many grams of NaCl are needed to make 500 ml of a 0.2 M solution? (MV=g/mol.wt.)
x  
(0.2 M) (0.5 L) = 

58.44  
Therefore:  
x  
0.1 = 

58.44  
= 5.844 g 
Important:
DO NOT CONFUSE THESE !!! 
If you want, you can use a formula triangle to work out molarity calculations:

e.g:
What is the concentration of a solution containing 4 moles of
solute in 2 L of solvent?
4 mol / 2 L = 2 M 
How many moles of solute are contained in 2 L of a 0.5 M solution?
2 L x 0.5 M = 1 mole 
What volume of 0.5 M solution can you make with 4 moles of solute?
4 mol / 0.5 M = 8 L 
Want to practice your molarity calculations?
Problems are randomly generated when you click the "New Problem" button.
Enter your answer in the empty square and click "Check Answer". The
results are displayed in the second table which will tell you whether you got
the correct answer or not and keeps a running total of your score. If you get
a question wrong, you can reenter and recheck your answer. Get a new problems
by clicking the "New Problem" button at any time.
Courtesy of George Wiger http://chemistry.csudh.edu
Dilutions of stock solutions are frequently used to make solutions of any desired molarity.
To dilute a solution means to add more solvent without the addition of more solute. (The resulting solution must be thoroughly mixed to ensure homogeneity.)
The fact that the solute amount stays constant allows calculations to be made:
moles before dilution = moles after dilution
From the definition of molarity (above):
moles of solute = molarity * volume
so we can substitute MV (molarity * volume) into the above equation:
Learn this equation! 
"1" refers to the situation before dilution and the "2" refers to after dilution. Volumes need not be converted to litres  any old volume measurement is fine, so long as the same one is used on each side.
The concentrations of solutions or suspensions are quoted not in molarities but in other terms such as weight per volume (e.g. "12 grams per litre" or "53 mg mL^{1}"), or as percentages.
Example 8:
You have 53 ml of a 1.5 M solution of NaCl, but 0.8 M solution is needed.
How many ml of 0.8 M can you make?
1.5 * 53 = 0.8 (V_{2})  
79.5  
= V_{2}  
0.8  
= 99.38 ml 
Example 9:
You need 225 ml of 0.6 M NaOH solution and you have a 2.5 M stock solution.
How would you make up the solution?
2.5 (V_{1}) = 0.6 * 225  
135  
V_{1} =  
2.5  
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If you want, you can use one of two methods to perform these calculations:
The SeeSaw Method: Rearrange the equation to keep the seesaw balanced:
M_{1}V_{1 }= M_{2}V_{2} 
The Ratio Method: The ratio of concentrations/volumes before/after dilution depends on the dilution: 
A) 50 ml of a
solution are diluted to a volume of 100 ml. The concentration of
the diluted solution is 2 M. What was the concentration of the original
solution? 



B) What volume of
a 5 M solution is required to make 100 ml of a 2 M solution? 

(2 * 0.1)/5 = 0.04 L 

C) 500 ml of a 5 M
solution are diluted to 2 L. What is the concentration of the resulting
solution? 

(5 * 0.5)/2 = 1.25 M 

D) 200 ml of a 1 M
solution are diluted to make a 0.4 M solution. What is the volume of the
resulting solution? 

(1 * 0.2)/0.4 = 0.5 L 

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