Answer:
P(C₆H₆) = 0.2961 atm
Explanation:
I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:
<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>
<em />
Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:
P = Xₐ * P° (1)
Where:
P: Partial pressure
Xₐ: molar fraction
P°: Vapour pressure
We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.
To get the moles: n = m / MM
To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:
MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol
MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol
Let's determine the moles of each compound:
moles (C₆H₆) = 79.2 / 78 = 1.02 moles
moles (C₇H₁₆) = 115 / 100 = 1.15 moles
moles in solution = 1.02 + 1.15 = 2.17 moles
To get the molar fractions, we use the following expression:
Xₐ = moles(C₆H₆) / moles in solution
Xₐ = 1.02 / 2.17 = 0.47
Finally, the partial pressure is:
P(C₆H₆) = 0.47 * 0.63
<h2>
P(C₆H₆) = 0.2961 atm</h2>
Hope this helps
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liquefies at
. About 1% of the air is still gas composed of the noble gas Argon, Ar.
