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Chapter 12: Problem 157

How many moles of potassium permanganate, \(\mathrm{KMnO}_{4}\), are there in \(28.86 \mathrm{~mL}\) of a \(5.20 \times 10^{-3} \mathrm{M}\) solution of \(\mathrm{KMnO}_{4} ?\)

Short Answer

Expert verified
There are approximately \(1.4992 \times 10^{-4}\) moles of potassium permanganate in the \(28.86 \mathrm{~mL}\) of the \(5.20 \times 10^{-3} \mathrm{M}\) solution.

Step by step solution

01

Convert the volume to liters

As 1 L = 1000 mL, we can convert the volume from mL to L. Divide 28.86 mL by 1000 to get the volume in liters: Volume (L) = 28.86 mL / 1000 Volume (L) = 0.02886 L
02

Calculate the number of moles

Now, we can use the formula to find the number of moles of potassium permanganate (KMnO4): Moles = Molarity × Volume Moles of KMnO4 = (5.20 × 10^(-3) M) × (0.02886 L) Moles of KMnO4 = 1.4992 × 10^(-4) Therefore, there are approximately \(1.4992 \times 10^{-4}\) moles of potassium permanganate in the 28.86 mL of the 5.20 x 10^-3 M solution.

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