The solubility of the oxidizing agent potassium permanganate is 7.1 g per 100.0 g of water at 25°C. What is the mole fraction of potassium permanganate (KMnO₄) in this solution? A) 0.0080 B) 0.0086 C) 0.066 D) 0.071 E) 0.48

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Quizplus · Accepted Answer

The mole fraction of potassium permanganate can be calculated using the formula:Mole fraction of KMnO4 = moles of KMnO4 / total moles of solutionFirst, we need to convert the solubility of KMnO4 to moles:Moles of KMnO4 = 7.1 g / 158.034 g/mol = 0.045 molNext, we need to calculate the total moles of solution. We can assume that the density of the solution is 1 g/mL, so 100.0 g of water is equal to 100.0 mL of solution. The molar mass of water is 18.015 g/mol, so the number of moles of water in the solution is:Moles of water = 100.0 g / 18.015 g/mol = 5.55 molFinally, we can calculate the mole fraction of KMnO4:Mole fraction of KMnO4 = 0.045 mol / (0.045 mol + 5.55 mol) = 0.0080Therefore, the correct answer is A) 0.0080.

The Solubility of the Oxidizing Agent Potassium Permanganate Is 7