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(158 g = 1 mole) --- molar mass of potassium permanganate.

You also need to specify the volume to be made. For 1 liter just add 15.8 g in a volumetric flask to make 1000 ml (1 liter) of solution.

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Q: How many grams KMnO4 take in 0.1N KMnO4 solution?
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You have .2 n kmno4 which you have to dilute it to 0.05n kmno4 how can you dilute that?

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This is more of a math question that requires a bit of knowledge of chemistry. So it helps to know the steps of this answer mathematically. Additionally it's worth noting that there are a number of ways to answer this question. The method I provide may take an extra step, but it allows for a better understanding of the process.First we need to know some basic information about potassium permanganate, KMnO4. This basic information can be found on a periodic table, like the one in the link below. The first step is finding the weight of oxygen in one mole of potassium permanganate as a percent. For this you need to know the atomic weights of the elements involved.K: 39.1 gramsMn: 54.9 gramsO: 16.0 grams × 4 atoms = 64.0 gramsKMnO4: 39.1 + 54.9 + 64.0 = 158.0 grams/molSo now we know the weight of one mole of potassium permanganate (158.0 grams). Because we also know the weight of oxygen, we can find the percent of oxygen in the compound by mass.64.0 grams O ÷ 158.0 grams KMnO4 = 0.405 = 40.5%In one mole of potassium permanganate, 64.0 grams of it is oxygen, meaning 40.5% of it is oxygen. Because of the Law of Definite Proportions, we know that in any amount of potassium permanganate, 40.5% of it is oxygen.Then you can set up an equation.40.5% of (some amount of KMnO4) = 27.5 grams oxygenLet's set the amount of KMnO4 as the variable "x".0.405x = 27.5x = 67.9 grams KMnO4


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