The vapor pressure of pure benzene at 25^circ C is 0.1252 atm. A solution prepared by dissolving 1.6 mol naphthalene (nonvolatile nondissociating) in 12 mol benzene will have a vapor pressure of __ atm. A) 0.110 B) 0.130 C) 0.080 D) 0.240 E) 0.060

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.<br /><br />Given information:<br />- Vapor pressure of pure benzene at 25°C = 0.1252 atm<br />- 1.6 mol naphthalene (nonvolatile nondissociating) is dissolved in 12 mol benzene<br /><br />Step 1: Calculate the total number of moles in the solution.<br />Total moles = Moles of naphthalene + Moles of benzene<br />Total moles = 1.6 mol + 12 mol = 13.6 mol<br /><br />Step 2: Calculate the mole fraction of benzene in the solution.<br />Mole fraction of benzene = Moles of benzene / Total moles<br />Mole fraction of benzene = 12 mol / 13.6 mol = 0.8824<br /><br />Step 3: Calculate the vapor pressure of the solution using Raoult's Law.<br />Vapor pressure of the solution = Vapor pressure of pure benzene × Mole fraction of benzene<br />Vapor pressure of the solution = 0.1252 atm × 0.8824 = 0.1104 atm<br /><br />Therefore, the vapor pressure of the solution is approximately 0.110 atm.<br /><br />The correct answer is A) 0.110.