Question

At a certain temperature the vapor pressure of pure benzene (C₆H₆) is measured to be 247 atm. Suppose a solution is prepared by mixing 131 g of benzene and 108. g of heptane (C₇,H₁₆). Calculate the partial pressure of benzene vapor above this solution.

Answer 1

To calculate the partial pressure of benzene vapor above the solution, we can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure.

Step-by-step explanation:

To calculate the partial pressure of benzene vapor above the solution, we can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure.

1. First, we need to calculate the mole fraction of benzene in the solution. The total mass of the solution is 131 g + 108 g = 239 g. The mole fraction of benzene is the moles of benzene divided by the total moles of both benzene and heptane. The moles of benzene can be calculated by dividing its mass by its molar mass: moles of benzene = 131 g / 78.11 g/mol = 1.678 mol. The total moles of the solution can be calculated by dividing the total mass by the molar mass of the solution: total moles of the solution = 239 g / (78.11 g/mol + 116.23 g/mol) = 1.678 mol + 2.042 mol = 3.72 mol. Therefore, the mole fraction of benzene is 1.678 mol / 3.72 mol = 0.451.
2. Next, we can calculate the partial pressure of benzene vapor above the solution using Raoult's law. The vapor pressure of pure benzene is 247 atm, so the partial pressure of benzene vapor above the solution is 0.451 * 247 atm = 111.54 atm.