Dilutions: Principles and Applications

Dilutions: Principles and Applications

Dilutions: Principles and Applications

Because solutions in science are often much more concentrated than are desired or can be managed for a given protocol, it is frequently necessary to dilute these solutions to a desired level. This requires a working knowledge of the principles of diluting, dilution factors, concentration factors and the calculations involved. High dilutions are usually expressed exponentially (i.e: a solution which has been diluted a million fold is termed a 106 dilution, or is 10-6 concentration).

Aliquot: a measured sub-volume of original sample.

Diluent: material with which the sample is diluted

Dilution factor (DF): ratio of final volume/aliquot volume (final volume = aliquot + diluent)

Concentration factor (CF): ratio of aliquot volume divided by the final volume (inverse of the dilution factor)
To calculate a dilution factor:

Remember that the dilution factor is the final volume/aliquot volume.

EXAMPLE: What is the dilution factor if you add 0.1 mL aliquot of a specimen to 9.9 mL of diluent?

  1.  The final volume is equal the the aliquot volume plus the diluent volume:  0.1 mL + 9.9 mL = 10 mL
  2. The dilution factor is equal to the final volume divided by the aliquot volume: 10 mL/0.1 mL = 1:100 dilution (10 2)

The Concentration Factor for this problem = aliquot volume/final volume = 0.1/(0.1 + 9.9) = 0.01 or 10 -2 concentration
To prepare a desired volume of solution of a given dilution:

  1. Calculate the volume of the aliquot: it is equal either to
  • the final volume/dilution factor


  • the concentration factor x final volume
  1. Calculate the volume of the diluent: which is equal to (the final volume – aliquot volume)
  2. Measure out the correct volume of diluent, add the correct volume of aliquot to it, mix.

EXAMPLE:  How would you prepare 20 mL of a 1:50 dilution?

  1. Determine required aliquot by dividing final volume by dilution factor:  20 mL/50 = 0.4 mL sample
  2. Subtract the aliquot volume from the final volume:  20 mL – 0.4 mL = 19.6 mL diluent
  3. Measure out 19.4 mL diluent, add 0.4 mL sample to it, mix thoroughly

SAMPLE PROBLEMS: (Note that these are in a different order than originally given out in class)

  1. How much sample (what sized aliquot) is required to prepare 10 mL of a 1 to 10 dilution, and how much diluent would you need?
  2. What is the dilution factor when 0.2 mL is added to 3.8 mL diluent? What is the concentration factor?
  3. You are to prepare 5 mL of a 102 dilution.  What should the aliquot and diluent volumes be?
  4. How would you prepare 20 mL of a 1:400 dilution?
  5. What is the dilution factor when you add 2 mL sample to 8 mL diluent?
  6. You add a pint of STP gas treatment to a 12 gallon fuel tank, and fill it up with gas. What is the dilution factor? (8 pints/gallon)
  7. You want 1 liter of 0.1 M NaCl, and you have 4 M stock solution. How much of the 4 M solution and how much dH2O will you measure out for this dilution?

For problems like the following, you need to know the ratio of the diluent to the aliquot.  For instance, if you are making a 1:20 dilution, the ratio of diluent to aliquot will be1 less than the dilution factor, or 19 parts diluent, 1 part aliquot:

  1. You have 0.6 mL of sample, and want to dilute it all to a fiftieth of its present concentration. How much diluent will you add, and what will the final volume be?
  2. You diluted a bacterial culture 106, plated out 0.2 mL and got 45 colonies on the plate. How many bacteria/mL were in the original undiluted culture?

A harder one which requires a little algebra:

  1. You have 100.00 mL of dH2O. How much glycerine would you have to add in order to make a 2.000 % v/v dilution?
    (Hint, let the volume of glycerine = X, set up the standard equation for a dilution factor using X, and solve for X.


1)  1 mL sample + 9.0 mL diluent

2)  DF = 20, CF = 0.05

3) aliquot = 0.05 mL, diluent = 4.95 mL

4) 0.05 mL sample, 19.95 mL diluent

5)  DF = 5

6) F.V. = 12 gallons x 8 pints/gallon = 96 pints.  Therefore 96pints/1 pint = D.F. =96

7) 25 mL 4.0 M stock solution + 975 mL dH2O

8) 29.4 mL diluent, final volume = 30 mL

9) 2.25 x 10^8

10)  ans: 2.04 mL glycerine  [ To solve:  a)  X/(100+X) = 0.02;   b)  X = 0.02 (100 +X);   c)  X = 2 + 0.02X ;  d)  X – 0.02X =2;  e)  0.98X =2;   f)  X = 2.04 ]

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