24) The vapor pressure of pure ethanol at 60^circ C is 46.508 kPa Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile)in 90.0 mmol ethanol will have a vapor pressure of __ kPa. A) 50.460 B) 41.847 C) 80.047 D) 37.186 E) 9.302 Answer:B

Question

24) The vapor pressure of pure ethanol at 60^circ C is 46.508 kPa Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile)in 90.0 mmol ethanol will have a vapor pressure of __ kPa. A) 50.460 B) 41.847 C) 80.047 D) 37.186 E) 9.302 Answer:B

Answer 1

To solve this problem, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.Given information:- Vapor pressure of pure ethanol at 60°C: 46.508 kPa- Amount of naphthalene (nonvolatile): 10.0 mmol- Amount of ethanol: 90.0 mmolStep 1: Calculate the total amount of moles in the solution.Total moles = Moles of naphthalene + Moles of ethanolTotal moles = 10.0 mmol + 90.0 mmol = 100.0 mmolStep 2: Calculate the mole fraction of ethanol in the solution.Mole fraction of ethanol = Moles of ethanol / Total molesMole fraction of ethanol = 90.0 mmol / 100.0 mmol = 0.9Step 3: Apply Raoult's Law to calculate the vapor pressure of the solution.Vapor pressure of the solution = Vapor pressure of pure ethanol × Mole fraction of ethanolVapor pressure of the solution = 46.508 kPa × 0.9 = 41.847 kPaTherefore, the correct answer is B) 41.847 kPa.

Soru

24) The vapor pressure of pure ethanol at 60^circ C is 46.508 kPa Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile)in 90.0 mmol ethanol will have a vapor pressure of __ kPa. A) 50.460 B) 41.847 C) 80.047 D) 37.186 E) 9.302 Answer:B

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