# Question #ec7e1

##### 1 Answer

Here's how you can do that.

#### Explanation:

All you have to do here is use the given *weight by volume percent concentration*, **conversion factor** to help you determine how many milliliters of solution would contain

A solution's weight by volume percent concentration tells you how many **grams of solute** you get **for every** **of solution**.

In your case, the solution is said to be

Now, the problem tells you that you have a mass of solute of *milligrams* to *grams* by using

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 g" = 10^3"mg")color(white)(a/a)|)))#

You will have

#2.5 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = "0.0025 g"#

So, you know that the solution contains *conversion factor* to find the volume of the solution that would contain

#0.0025 color(red)(cancel(color(black)("g solute"))) * overbrace("100 mL solution"/(0.80color(red)(cancel(color(black)("g solute")))))^(color(blue)("= 0.80% w/v")) = "0.3125 mL solution"#

Rounded to two **sig figs**, the answer will be

#"volume of solution" = color(green)(|bar(ul(color(white)(a/a)"0.31 mL"color(white)(a/a)|)))#