How to prepare 1 liter of 0.05 N potassium permanganate solution from chemically pure KMnO4?
1 – The normal concentration is determined by the formula
Cn = n * e / V, where n is the number of moles of the substance, e is the equivalent of the substance (for KMnO4 it is 1), V is the volume of the solution.
Fromsuda n = Cn * V / e
n = 0.05n * 1 l / 1
n = 0.05 mol.
2 – The mass of a substance is determined by the formula m = n * M, where M is the molar mass of potassium permanganate. (behind the periodic table M KMnO4 = 39 + 55 + 16 * 4 = 158 g / mol.
m = 0.05 mol * 158 g / mol = 7.9 g.
3 – The mass of water required to prepare the solution is
1000 g – 7.9 g = 992.1 g, since the volume of water 1 liter is equal to the mass of 1 kg or 1000 g.
Answer: 7.9 grams of pure KMnO4 and 992.1 grams of water are needed to prepare 1 liter of 0.05 n KMnO4 solution