# Question #3e8e7

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The idea here is that when you're performing a **dilution**, you can equate the ratio that exists between the volume of the diluted solution and the volume of the stock solution with the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution.

For any dilution, you will have

#"volume diluted"/"volume stock" = "concentration stock"/"concentration diluted"#

In your case, the stock solution must be diluted by a factor of

#"DF" = (60color(red)(cancel(color(black)(%))))/(6color(red)(cancel(color(black)(%)))) = color(blue)(10) -># this is thedilution factor

This means that the volume of the stock solution

#"volume diluted"/"volume stock" = color(blue)(10)#

will be equal to

#"volume stock" = "volume diluted"/color(blue)(10)#

Plug in your value to find

#"volume stock" = "300 mL"/color(blue)(10) = "30 mL"#

You can thus say that the volume of diluent needed to make your solution will be

#"volume diluent" = overbrace("300 mL")^(color(red)("volume diluted solution")) - overbrace("30 mL")^(color(red)("volume stock solution"))#

#color(darkgreen)(ul(color(black)("volume diluent = 270 mL")))#

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

So, in order to prepare your solution, you need to take *enough water*, which, for all intended purposes, will be equal to