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Heptane, C7H16, and octane, C8H18, form ideal solution. The solution boils at 85.00 ºC and 0.522 atm. (At 85.00 ºC, the vapor pressure of pure heptane is 0.570 atm and the vapor pressure of pure octane is 0.234 atm.)
What is the vapor pressure above the solution if the mole fraction of octane is 0.234?
To find the vapor pressure above the solution, we will use Raoult's Law, which states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.
First, let's calculate the mole fraction of heptane. Since the mole fraction of octane is given as 0.234, we can find the mole fraction of heptane as follows:
Mole fraction of heptane = 1 - Mole fraction of octane
Mole fraction of heptane = 1 - 0.234
Mole fraction of heptane = 0.766
Next, we will use Raoult's Law to find the vapor pressure above the solution. According to Raoult's Law, the vapor pressure above the solution is equal to the sum of the vapor pressure of each component multiplied by its mole fraction.
Vapor pressure above the solution = (Mole fraction of heptane × Vapor pressure of heptane) + (Mole fraction of octane × Vapor pressure of octane)
Vapor pressure above the solution = (0.766 × 0.570 atm) + (0.234 × 0.234 atm)
Calculating this expression, we find:
Vapor pressure above the solution = 0.436 atm + 0.055 atm
Vapor pressure above the solution = 0.491 atm
Therefore, the vapor pressure above the solution, when the mole fraction of octane is 0.234, is 0.491 atm.