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I will assume that you will start from the crystals of permanganate:
Calculations:
M.M. potassium permanganate: 158.04 g/mol
mol KMnO4 in 10mL sol'n: 1.5 mol/L x 10 mL x (1 L / 1000 mL) = 0.015 mol
grams potassium permanganate: 0.015 mol x 158.04 g/mol = 2.3706 g / 10 mL sol'n
Preparation:
1. Weigh out analytically 2.3706g KMnO4 into a 10 mL volumetric flask.
2. Dilute to the mark with dH2O.
To prepare a 100 ppm (parts per million) solution of KMnO4, you would dissolve 0.1 grams of KMnO4 in 1 liter of water. This concentration means there are 100 milligrams of KMnO4 dissolved in 1 liter of solution.
First we look at the equation involving KMnO4 in Redox reaction.
MnO4- + 8H+ + 5e- -------> Mn2+ + 4H2O
Hence 5 electrons are transferred in the reaction. Therefore 1N solution of KMnO4 = M/5 solution of KMnO4 i.e., 0.20M KMnO4 solution. Hence 0.1N KMNO4 solution is equivalent to 0.02M KMnO4 i.e., a solution containing 3.1607g/L of KMnO4
1. Weigh 15,803 g ultrapure KMnO4 dried at 110 0C for 30 min.
2. Transfer KMnO4 in a clean 1 L volumetric flask using a funnel. 3. Wash the funnel with 0,9 L demineralized water.
4. Put the flask in a thermostat and maintain 30 min at 20 0C.
5. Add demineralized water up to the mark.
6. Stir vigorously and transfer in a clean bottle with stopper.
7. Add a label with necessary information.
The molar mass of potassium permanganate is 158,034; 2 M KMnO4 equals 316,068 g.
1000 mL----------------------------316,068 g
20 mL-------------------------------x
x= 20 x 316,068/1000 = 6,32 g
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Preparation: Dissolve 6,32 g KMnO4 a.r. dried in 20 mL demineralized water, at 20 oC, in a volumetric flask.
there is tow method to prepare 1n kmno4 it's depend on oxidation number. Iknow one of this method you need 31.6 gms in one L D.WATER IF YOU KNOW FURTHER INFORMATION PLEASE TELL ME.
By solving 100 mg of KMnO4 in 1 L of distilled water at 20 0C.
Dissolve 1,58 g KMnO4 for analysis in 1 L distilled water at 20 oC.
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