# Question 18314

Jun 20, 2017

The molarity of the ${\text{KMnO}}_{4}$ solution is $\text{0.01511 mol/L}$. A container of this solution would be labeled $\text{0.01511 M KMnO"_4}$, and it would be called a $0.01511$ molar potassium permanganate solution.

#### Explanation:

Molarity $= \text{mol solute"/"L of solution}$.

You have liters of solution, but you don't have moles of solute, $\text{KMnO"_4}$ (potassium permanganate).

However, you can determine moles ${\text{KMnO}}_{4}$ from its given mass and its molar mass, which is $\text{158.0323 g/mol}$.

To determine the moles of $\text{KMnO"_4}$, multiply its given mass by the inverse of its molar mass.

4.7744color(red)cancel(color(black)("g KMnO"_4))xx(1"mol KMnO"_4)/(158.0323color(red)cancel(color(black)("g KMnO"_3)))="0.030212 g KMnO"_4"#

Now that you have moles of solute and liters of solution, you can calculate molarity using the equation above.

Molarity$= \left(0.030212 \text{mol KMnO"_4)/(2.000"L solution")=("0.01511 mol KMnO"_4)/(1"L solution}\right)$