Chapter 9 Source Answers and Examples
Chapter problems → · Original reading →
Reviewed solution to Problem 5
New reviewed response for this exact prompt.
A polarOH group allows butanol–water hydrogen bonding, while its carbon chain also affects dissolution. Methane and octane are nonpolar, but solubility depends on more than a polarity label. SO₂ participates in aqueous equilibria, so physical dissolution and reaction both matter. A full rank needs measured solubilities at a common temperature and defined pressure/speciation conditions; the supplied labels alone do not certify one.
Reviewed solution to Problem 8
New reviewed response for this exact prompt.
OnlyX partial pressure determines its saturation. ΔP=0.150−0.100=0.050atm. Δn=(0.050)(1.00)/(0.082057×300.0)=0.0020311mol before final rounding. The pressure difference has two significant figures, so report0.0020mol. Total inert-gas pressure does not replaceX partial pressure. The explicit new model replaces the under-specified atmospheric benzene story.
Reviewed solution to Problem 13
New reviewed response for this exact prompt.
UsePi=xiPtotal. A:PO₂=0.210atm,PN₂=0.790atm. B:PO₂=0.300atm,PN₂=0.450atm. Components sum to their sample’s stated total. These are supplied hypothetical conditions, not a universal airline-cabin composition/pressure claim.
Reviewed solution to Problem 15
New reviewed response for this exact prompt.
Boiling occurs whenPexternal=P°X+P°Y=0.1000atm. Vapor fractions arexX=0.0800/0.1000=0.800andxY=0.200. Both liquids contribute vapor; independent normal-boiling-point stages are not the model. The newly suppliedX/Y data replace the original under-specified benzene/water request.
Reviewed solution to Problem 17
New reviewed response for this exact prompt.
5.00°C=278.15K, aboveTc190.6K. There is no separate liquid–gas phase boundary above the critical temperature, so compression alone does not produce the requested distinct liquid methane phase. Extrapolating that Antoine fit beyond its stated range is invalid.
Worked examples and reviewed revisions
Example Problem 1.1.1
Original example and steps
Example Problem 1.1.1
Determine whether H2O is Polar or Non-Polar, and identify any partial charges on the molecule.
Step 1: Draw out the Molecule
In this case our molecule is water (H2O). This is a simple bent molecule. Therefore we can draw it as such:
Step 2: Identify points of high and low electronegativity
We know that Hydrogen will have a low electronegativity, and Oxygen will be pretty high.
Step 3: Place the partial charges (if any)
We will have partial charges, as we have a polar molecule. We know that oxygen represents the most highly electronegative portion of the atom, so we can therefore place our -δ there. As for the +δ, we have two places left: both Hydrogen atoms. As each atom has the same electronegativity, we will need to draw two positive partial charges. This gives us the following:
Step 4: Determine the polarity of the Molecule
As we have a large discrepancy of electronegativity throughout the molecule, we can say that the molecule is going to be polar.
Example Problem 1.1.2
Original example and steps
Example Problem 1.1.2
Determine whether O2 is Polar or Non-Polar, and identify any partial charges on the molecule.
Step 1: Draw out the Molecule
Whenever you have only 2 atoms bonded, it always results in a linear structure:
Step 2: Identify points of high and low electronegativity
We recall that Oxygen is very electronegative, however as this molecule only contains Oxygen, there is no relative point which would be considered more electronegative.
Step 3: Place the partial charges (if any)
As the molecule has no points of high or low electronegativity, there are no partial charges.
Step 4: Determine the polarity of the Molecule
Due to the uniformity of the electronegativity, we can say that this molecule is non-polar. This might seem weird as Oxygen is so stereotypically electronegative, but due to the uniformity of the molecule, we resolve to say that it is non-polar.
Example Problem 1.1.3
Original example and steps
Determine whether C6H14O is Polar or Non-Polar, and identify any partial charges on the molecule.
Step 1: Draw out the Molecule
This molecule is pretty lengthy, so we have the following structure:
Step 2: Identify points of high and low electronegativity
We decided previously that Carbon and Hydrogen are both have similarly low electronegativity. However, that Oxygen on the end has a high electronegativity.
Step 3: Place the partial charges (if any)
This molecule has quite a few bonds, but the local dipoles in most of the carbon-hydrogen bonds would cancel one another out. To only look at the relevant partial charges, we will leave those out. This leaves us with:
Step 4: Determine the polarity of the Molecule
Thanks to the presence of the oxygen, this molecule is Polar.
Example Problem 1.3.1
Worked example
AP edition reviewed replacement.
Vapor-pressure enrichment: estimate water’s boiling pressure at35.00°C using NIST’s fitA5.20389,B1733.926,C−39.485, valid304–333K, with log₁₀(P/bar)=A−B/(T/K+C).
New independently computed worked answer.
T=308.15K. UsingP=10^[5.20389−1733.926/(308.15−39.485)], the pressure is0.056238bar or0.055503atm to five significant figures. At equilibrium boiling this vapor pressure matches the specified external pressure. The old coefficients and0.0603atm result are superseded.