# Question #cec3a

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you're performing a **serial dilution**, so you should be aware of the fact that the **overall dilution factor** will be equal to the *product* of the dilution factors of each individual dilution.

#"DF"_"overall" = "DF"_1 xx "DF"_2 xx ... xx "DF"_n#

As you know, the **dilution factor** can be calculated by dividing the volume of the *diluted solution* by the volume of the *concentrated solution*.

#"DF" = V_"diluted"/V_"concentrated"#

In your case, you're performing **identical dilutions** that have

#"DF" = ((1 + 9)color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("mL")))) = 10#

That is the case because, for each dilution, the volume of the concentrated solution is equal to

So, you can say that after **dilutions**, the overall dilution factor will be

#"DF"_"8 dilutions" = overbrace(10 xx 10 xx... xx 10)^(color(blue)("8 times")) = 10^8#

Now, the dilution factor also tells you the ratio that exists between the concentration of the *concentrated solution* and the concentration of the *diluted solution*.

For the **overall dilution**, you have

#"DF"_ "8 dilutions" = c_"initial"/c_"final"#

This means that the final concentration of the solution will be

#c_"final" = c_"initial"/10^8#

In your case, this is equivalent to

#c_"final" = "0.1 M"/10^8 = 10^(-9)# #"M"#

Now, *for all intended purposes* and based on the number of significant figures that you have for your values, you can go ahead and say that the

It's worth mentioning that the actual