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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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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

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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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 of naphthalene is dissolved in 0.22 mol of 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 + 0.22 mol = 1.82 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 = 0.22 mol / 1.82 mol = 0.1213<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 × 0.1213 = 0.0152 atm<br /><br />Therefore, 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.<br /><br />The correct answer is E) 0.060.
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