The vapor pressure of pure benzene at 25^circ C is 0.1252 atm. A.solution prepared by dissolving 1.6 mol naphthalene Inomolatile, nondisocialingles.22 mailbenzonewill have a vapor pressure of __ atm. A) 0.110 B) 0.130 C) 0.080 D) 0.24 E) 0.060

Question

The vapor pressure of pure benzene at 25^circ C is 0.1252 atm. A.solution prepared by dissolving 1.6 mol naphthalene Inomolatile, nondisocialingles.22 mailbenzonewill have a vapor pressure of __ atm. A) 0.110 B) 0.130 C) 0.080 D) 0.24 E) 0.060

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 benzene at 25°C = 0.1252 atm- 1.6 mol of naphthalene is dissolved in 0.22 mol of benzene.Step 1: Calculate the total number of moles in the solution.Total moles = Moles of naphthalene + Moles of benzeneTotal moles = 1.6 mol + 0.22 mol = 1.82 molStep 2: Calculate the mole fraction of benzene in the solution.Mole fraction of benzene = Moles of benzene / Total molesMole fraction of benzene = 0.22 mol / 1.82 mol = 0.1213Step 3: Calculate the vapor pressure of the solution using Raoult's Law.Vapor pressure of the solution = Vapor pressure of pure benzene × Mole fraction of benzeneVapor pressure of the solution = 0.1252 atm × 0.1213 = 0.0152 atmTherefore, the vapor pressure of the solution prepared by dissolving 1.6 mol of naphthalene in 0.22 mol of benzene will be 0.0152 atm.The correct answer is E) 0.060.

Soru

The vapor pressure of pure benzene at 25^circ C is 0.1252 atm. A.solution prepared by dissolving 1.6 mol naphthalene Inomolatile, nondisocialingles.22 mailbenzonewill have a vapor pressure of __ atm. A) 0.110 B) 0.130 C) 0.080 D) 0.24 E) 0.060

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