# Question 269be

Feb 5, 2015

The dilution factor for the fourth tube will be $\text{1:160,000}$.

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 $\text{0.2 mL}$ to $\text{3.8 mL}$, 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

$\text{DF" = V_("final")/V_("initial") = "4.0 mL"/"0.2 mL" = "20}$, or $\text{1:20}$

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 $\text{1:20}$ dilution factor. Now you take $\text{0.2 mL}$ of this solution and add it to another $\text{3.8 mL}$. The second tube will have a dilution factor equal to $\text{1:20}$, but the original solution has now been diluted twice by a dilution factor of $\text{1:20}$, 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 $\text{1:160,000}$.

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