MicrobiologyBytes: Maths & Computers for Biologists: Molarities & Dilutions Updated: October 19, 2004 Search

Molarities and Dilutions

Further information on this topic can be found in Chapter 4 of:

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A mole!

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:

"the amount of a substance which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12"

Complicated, isn't it?

Amedeo Avogadro 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:

equal volumes of all gases at the same temperature and pressure contain the same number of molecules

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:

602,000,000,000,000,000,000,000   (6.02*1023)

How big is Avogadro's number?

  • An Avogadro's number of soft drink cans would cover the surface of the earth to a depth of over 200 miles.
  • If you spread Avogadro's number of unpopped popcorn kernels across the USA, the entire country would be covered in popcorn to a depth of over 9 miles.
  • If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

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:

  moles of solute
 Molarity  =  
  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:

  1. Convert grams to moles
  2. Divide moles by litres to calculate molarity
a) The molecular weight of NaOH is 40 grams/mol

40 grams  

  =  1 mol
40 grams / mol  

b)

1 mol  

  =  0.5 M
2 L  

 

Learn these equations!

 

grams  

  =  moles (mol) 
 molecular weight  

 

moles  

  =  molarity (M) 
 volume  

 

  grams
 Molarity (M) * Volume (V)  =  
  molecular weight 

 

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:

  1. Measurement in moles is a measurement of the amount of a substance.

  2. Measurement in molarity is a measurement of the concentration of a substance - the amount (moles) per unit volume (litres).

DO NOT CONFUSE THESE !!!

 

Still not confident about molarities?

If you want, you can use a formula triangle to work out molarity calculations:

  mol
 M  =  
 
formula triangle

e.g:

What is the concentration of a solution containing 4 moles of solute in 2 L of solvent?

formula triangle 4 mol / 2 L = 2 M

 

How many moles of solute are contained in 2 L of a 0.5 M solution?

formula triangle 2 L x 0.5 M = 1 mole

 

What volume of 0.5 M solution can you make with 4 moles of solute?

formula triangle 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 re-enter and recheck your answer. Get a new problems by clicking the "New Problem" button at any time.

Moles Vol(L) Molarity
Results Total Done Total Correct

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:

 M1V1  =  M2V2 

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 (V2)
 
79.5  

  =  V2
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?

=54 ml
2.5 (V1)  =  0.6 * 225
 
  135
V1  =  
  2.5
 

 

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Still not confident about dilutions?

If you want, you can use one of two methods to perform these calculations:

The See-Saw Method:

Rearrange the equation to keep the see-saw balanced:

equation

M1V= M2V2 

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?

equation


(2 * 0.1)/0.05 = 4 M

  • 50 ml -> 100 ml = 2-fold dilution.
  • Concentration of original solution (what you are trying to calculate) is GREATER than concentration of diluted solution, so MULTIPLY by the ratio of the volumes:
  • 2 M * 2 = 4 M
B) What volume of a 5 M solution is required to make 100 ml of a 2 M solution?

equation

(2 * 0.1)/5 = 0.04 L
(= 40 ml)

  • 5 M -> 2 M = 2.5-fold dilution.
  • Volume of original solution (what you are trying to calculate) is LESS than volume of diluted solution, so DIVIDE by the ratio of the concentrations:
  • 100 / 2.5 = 40 ml
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

  • 0.5 L -> 2 L = 4-fold dilution.
  • Concentration of resulting solution (what you are trying to calculate) is LESS than concentration of original solution, so DIVIDE by the ratio of the volumes:
  • 5 M / 4 = 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?

equation

(1 * 0.2)/0.4 = 0.5 L
(= 500 ml)

  • 1 M -> 0.4 M = 2.5-fold dilution.
  • Volume of resulting solution (what you are trying to calculate) is GREATER than volume of original solution, so MULTIPLY by the ratio of the volumes:
  • 200 * 2.5 = 500 ml

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