Question

Question #2a6d3

Answer 1

`"4.7 dm"^3`

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)`, which basically means that you get `"3.17 g"` of potassium permanganate, the solute, for every `"1 dm"^3` of solution.

`"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 g"` of solute

`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.