What volume of 0.125 M KMnO4 is required to yield 0.180 mol of potassium permanganate, KMnO4
To find the volume of 0.125 M KMnO4 required to yield 0.180 mol of KMnO4, we first need to use the formula:
\[ \text{Volume} = \frac{\text{Amount}}{\text{Molarity}} \]
Given:
- Amount of KMnO4 = 0.180 mol
- Molarity of KMnO4 = 0.125 M
Substitute these values into the formula:
\[ \text{Volume} = \frac{0.180 \, \text{mol}}{0.125 \, \text{M}} \]
\[ \text{Volume} = 1.44 \, \text{L} \]
Therefore, 1.44 liters of 0.125 M KMnO4 is required to yield 0.180 mol of KMnO4.