What is the mole ratio of benzene (C6H6) to n-octane in the vapor above a solution of 15.0% benzene and 85.0% n-octane by mass at 25 degrees Celcius? the vapor pressures of n-octane and benzene are 11 torr and 95 torr.
here is what I did (check my work, because something isnt right):
15g C6H6/78.12gC6H6 = .19 mol C6H6
85g C8H18/114.26g C8H18= .74 mol C8H18
X-benzene: .19/.93= .204
X-octane: .74/.93= .796
pBen: .204*95=19.38
pOctane: .796*11=8.756
19.38+ 8.756= 28.136
15% benzene means 15g benzene/(15g benzene + 85 g octane).
15/molar mass benzene = mols benzene.
85/molar mass octane = mols octane.
Calculate Xbenzene
Calculate Xoctane.
Then pBen = XBen*Pobenzene
and pOct = XOct*Pooctane
Then Ptotal = Pbenzene + Poctane
That looks ok to me. You haven't been consistent with significant figures. If your database and prof are picky about that you need to go through and make some changes.
The database said the answer was 2.23 ??
That' because you and I didn't finish the question. Go back and look at the problem. It asks for the benzene/octane RATIO. So take pbenzene/poctane. Using your numbers that is 19.38/8.756 = 2.21 and I suspect the difference (2.21 vs 2.23) is one of significant figures through the computation.
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To determine the mole ratio of benzene (C6H6) to n-octane in the vapor above the solution, we need to consider the vapor pressure of each component and their relative amounts in the solution.
Here's how we can approach this problem step by step:
Step 1: Calculate the masses of benzene and n-octane in the solution.
- Assume we have 100 grams of the solution.
- The solution is 15.0% benzene by mass, so the mass of benzene can be calculated as:
Mass of benzene = 15.0 grams (100 grams)
- Similarly, the mass of n-octane can be calculated as:
Mass of n-octane = 85.0 grams (100 grams)
Step 2: Convert the masses of benzene and n-octane to moles.
- First, we need to calculate the molecular weight of benzene and n-octane.
Molecular weight of benzene (C6H6) = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
Molecular weight of n-octane (C8H18) = 8(12.01 g/mol) + 18(1.01 g/mol) = 114.23 g/mol
- Now, we can convert the masses to moles using the molecular weights:
Moles of benzene = Mass of benzene / Molecular weight of benzene
= 15.0 g / 78.11 g/mol
Moles of n-octane = Mass of n-octane / Molecular weight of n-octane
= 85.0 g / 114.23 g/mol
Step 3: Calculate the mole ratio of benzene to n-octane.
- The mole ratio of benzene to n-octane can be determined by dividing the moles of benzene by the moles of n-octane:
Mole ratio of benzene to n-octane = Moles of benzene / Moles of n-octane
For example, if we calculate the moles and mole ratio using the given masses and molecular weights, we would have:
Moles of benzene = 15.0 g / 78.11 g/mol ≈ 0.192 mol
Moles of n-octane = 85.0 g / 114.23 g/mol ≈ 0.743 mol
Mole ratio of benzene to n-octane = 0.192 mol / 0.743 mol ≈ 0.258
Therefore, the mole ratio of benzene to n-octane in the vapor above the solution is approximately 0.258.