# Question #2a6d3

##### 1 Answer

#### Answer:

#### Explanation:

Well, all you really have to do here is to use the known concentration of the solution as a *conversion factor* to go from mass of potassium permanganate to volume of solution.

You know that this solution has a concentration of

#"3.17 g dm"^(-3) = "3.17 g KMnO"_4/("1 dm"^3color(white)(.)"solution")#

In your case, you want to go from *grams* of potassium permanganate to *cubic decimeters* of solution, so rewrite the conversion factor as

#("1 dm"^3color(white)(.)"solution")/"3.17 g KMnO"_4#

Use this to calculate the volume of solution that contains

#15 color(red)(cancel(color(black)("g KMnO"_4))) * ("1 dm"^3color(white)(.)"solution")/(3.17color(red)(cancel(color(black)("g KMnO"_4)))) = color(darkgreen)(ul(color(black)("4.7 dm"^3color(white)(.)"solution")))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the mass of potassium permanganate.