Question

Question #18314

Answer 1

The molarity of the `"KMnO"_4` solution is `"0.01511 mol/L"`. A container of this solution would be labeled `"0.01511 M KMnO"_4"`, and it would be called a `0.01511` molar potassium permanganate solution.

Explanation:

Molarity `="mol solute"/"L of solution"`.

You have liters of solution, but you don't have moles of solute, `"KMnO"_4"` (potassium permanganate).

However, you can determine moles `"KMnO"_4` from its given mass and its molar mass, which is `"158.0323 g/mol"`.

To determine the moles of `"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`=(0.030212"mol KMnO"_4)/(2.000"L solution")=("0.01511 mol KMnO"_4)/(1"L solution")`