# Question #f3a0d

##### 1 Answer

Here's what I got.

#### Explanation:

Your target solution must be **for every** **of solution**.

Now, your starting solution is **for every** **of solution**.

In order to reduce the concentration of the solution from *dilute* the sample.

For any dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution will give you the **dilution factor**,

In this case, you have

#"DF" = (5 color(red)(cancel(color(black)(%))))/(2color(red)(cancel(color(black)(%)))) = color(blue)(2.5)#

The trick here is to realize that the **dilution factor** is also equal to the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

#"DF" = V_"diluted"/V_"stock"#

This means that

#V_"diluted" = "DF" * V_"stock"#

In your case, you have

#V_"diluted" = color(blue)(2.5) * "10 mL" = "25 mL"#

So if the volume of the *diluted solution* is equal to

#"25 mL " - " 10 mL" = "15 mL"#

of water. In other words, if you add

I'll leave the answer rounded to two **sig figs**, but keep in mind that your values justify an answer rounded to one significant figure.