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

Nov 21, 2016

$\text{36.8 mL}$

#### 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, ${\text{KMnO}}_{4}$, is said to have a molarity of ${\text{0.124 mol L}}^{- 1}$. This tells you that every $\text{1 L}$ of this solution will contain $0.124$ 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 ${\text{158.034 g mol}}^{- 1}$, which means that $0.124$ 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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 L" = 10^3"mL}}}}$

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

$\left(\text{0.124 moles KMnO"_4)/("1 L solution") = ("19.60 g KMnO"_4)/(10^3"mL solution}\right)$

Now you're ready to use this as a conversion factor to determine how many milliliters of solution would contain $\text{0.722 g}$ of potassium permanganate

$0.722 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g KMnO"_4))) * (10^3"mL solution")/(19.60color(red)(cancel(color(black)("g KMnO"_4)))) = color(darkgreen)(ul(color(black)("36.8 mL solution}}}}$

The answer is rounded to three sig figs.