Question

Question #269be

Answer 1

The dilution factor for the fourth tube will be `"1:160,000"`.

You know that the dilution factor is defined as the final volume of the sample divided by the initial volume of the sample. So, if you add `"0.2 mL"` to `"3.8 mL"`, you initial volume will be

`V_("initial") = "0.2 mL"`

and your final volume will be

`V_("final") = "0.2 + 3.8" = "4.0 mL"`

This means that the dilution factor will be

`"DF" = V_("final")/V_("initial") = "4.0 mL"/"0.2 mL" = "20"`, or `"1:20"`

When you're doing a serial dilution, each tube will dilute the sample by the same calculated dilution factor.

Notice how in the above example each solution will be diluted by a factor of 10. If you apply this method to your dilution, you'll get a dilution factor for tube 4 equal to

`DF_("final tube") = "1:20"^(4) = "1:160,000"`

The first tube will have a `"1:20"` dilution factor. Now you take `"0.2 mL"` of this solution and add it to another `"3.8 mL"`. The second tube will have a dilution factor equal to `"1:20"`, but the original solution has now been diluted twice by a dilution factor of `"1:20"`, which means that the solution in tube 2 will have a total dilution factor of

`DF_("tube 2") = "1:20" * "1:20" = "1:20"^(2) = 1:400`

Now you repeat the process for tubes 3 and 4 to get the final dilution factor of `"1:160,000"`.

You can read more on dilution factors and serial dilutions here:

http://socratic.org/questions/how-do-you-calculate-dilution-factor?source=search

http://socratic.org/questions/how-do-you-do-serial-dilution-calculations?source=search