# How many mL of a 0.124 moles/L solution of potassium permanganate contain 0.722 g of the salt?

##### 1 Answer

#### Explanation:

As you know, **molarity** is a measure of how many *moles of solute* you get **for every liter** of a given solution.

In your case, a solution of sodium permanganate, **molarity** of **every** **moles** of solute.

Now, one cool way of solving this problem would be to **convert** the molarity of the solution from *moles per liter* to *grams per milliliter*.

Potassium permanganate has a **molar mass** of **moles** of potassium permanganate would be equivalent to

#0.124 color(red)(cancel(color(black)("moles KMnO"_4))) * "158.034 g"/(1color(red)(cancel(color(black)("mole KMnO"_4)))) = "19.60 g"#

You also know that you have

#color(blue)(ul(color(black)("1 L" = 10^3"mL")))#

which means that the molarity of the solution is equivalent to a concentration of

#("0.124 moles KMnO"_4)/("1 L solution") = ("19.60 g KMnO"_4)/(10^3"mL solution")#

Now you're ready to use this as a **conversion factor** to determine how many *milliliters* of solution would contain

#0.722 color(red)(cancel(color(black)("g KMnO"_4))) * (10^3"mL solution")/(19.60color(red)(cancel(color(black)("g KMnO"_4)))) = color(darkgreen)(ul(color(black)("36.8 mL solution")))#

The answer is rounded to three **sig figs**.