The solubility of the oxidizing agent potassium permanganate is 7.1 g per 100.0 g of water at 25$\degree$C. What is the mole fraction of potassium permanganate in this solution?
A) 0.0080
B) 0.0086
C) 0.066
D) 0.45
E) 0.48

Question

Quizplus · Accepted Answer

The mole fraction of potassium permanganate is calculated by dividing the moles of potassium permanganate by the total moles of solute and solvent. The moles of potassium permanganate can be calculated by dividing the mass of potassium permanganate by its molar mass. The moles of water can be calculated by dividing the mass of water by its molar mass. The total moles of solute and solvent is the sum of the moles of potassium permanganate and the moles of water. The mole fraction of potassium permanganate is then calculated as follows:$$	ext{Mole fraction of potassium permanganate} = \frac{	ext{Moles of potassium permanganate}}{	ext{Total moles of solute and solvent}}$$$$=\frac{7.1 	ext{ g} / 158.034 	ext{ g/mol}}{7.1 	ext{ g} / 158.034 	ext{ g/mol} + 100.0 	ext{ g} / 18.015 	ext{ g/mol}}$$$$= 0.0080$$

The Solubility of the Oxidizing Agent Potassium Permanganate Is 7 °\degree°

The Solubility of the Oxidizing Agent Potassium Permanganate Is 7 $\degree$