# What volume of 0.125 M KMnO4 is required to yield 0.180 mol Of potassium permanganate KMnO4?

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To find the volume of 0.125 M KMnO4 required to yield 0.180 mol of potassium permanganate (KMnO4), you can use the formula:

[ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} ]

Given: Molarity (M) = 0.125 M Number of moles = 0.180 mol

Substituting the values into the formula:

[ \text{Volume} = \frac{0.180 , \text{mol}}{0.125 , \text{mol/L}} ]

[ \text{Volume} = \frac{0.180}{0.125} , \text{L} ]

[ \text{Volume} = 1.44 , \text{L} ]

So, 1.44 liters of 0.125 M KMnO4 is required to yield 0.180 mol of potassium permanganate (KMnO4).

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