A solution contains 50.0 g of heptane (C7H16) and 50.0g of octane (C8H20) at 25oC. The vapor pressures of pure heptane and pure octane at this temperature are 45.8 and 10.9 Torr, respectively.
Assuming ideal behavior, calculate the:
1) vapor pressure of each of the solution components in the mixture
2) total pressure above the solution
3) composition of the vapor in mass percent
Please explain! Thank you.
Answer: According to the question : You can calculate the total vapor pressure of the solution by multiplying the vapor pressure of each component by its mole fraction in the mixture, and then adding those together.
mole fraction of heptane = moles heptane / (moles heptane + moles octane)
moles of heptane = 50 / 100.21 = 0.49
moles of octane = 50/114.23 = 0.44
mole fraction of heptane = 0.49 / 0.49+ 0.44 = 0.526
Mole fraction of octane = 0.44 / 0.44+ 0.49 = 0.474
Now, we have to use Pa = Xa * Ptotal
Ptotal = Pa/Xa = 45.8/0.526 = 87.07 atm
For Octane = 10.9 / 0.474 = 22.995 atm
And the total pressure is = P1 + P2 = 87.07 + 22.995 = 110.065 atm
Hence it is all about the given question Thank you :)
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