TBR Chem #29 passage 5 Phase Changes - vapor pressure and density (p102, 2010ed)

[Gauss44]

2 Questions: 29 & 31

29. Above which beaker is the vapor pressure of methanol the greatest?

A. Beaker I
B. Beaker II
C. Beaker III
D. The vapor pressure of methanol is the same above all 3 beakers

Correct answer: B (highlight to show correct answer)

Passage Info.

All beakers are sealed.
Beaker 1 contains: 50g methanol and 50g ethanol
Beaker 2 contains: 50ml methanol and 50 ml ethanol
Beaker 3 contains: 1 mole methanol and 1 mole ethanol

Density of methanol: .7914g/ml (mw = 32; bp = 56 degrees C)
Density of ethanol: .7893g/ml (mw = 46; bp = 79 degrees C)

Answer Key: The solution with the greatest mole fraction of methanol has the highest vapor pressure of methanol. Methanol has the lower molecular mass between ethanol and methanol, so beaker 1 has more moles of methanol than ethanol. This means that beaker 1 will exhibit a higher vapor pressure due to methanol than beaker 3. Because the density of methanol is greater than the density of ethanol, beaker 2 with equal volume quantities of methanol and ethanol has more methanol by mass than ethanol by mass. The mole fraction of methanol in beaker 2 is even greater than the mole fraction of ethanol in beaker 1. The highest vapor pressure of methanol is therefore above beaker 2.

Can anyone calculate step by step?



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Question # 31 below
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31. During the first minute of evaporation, before equilibrium is established and the temperature changes to any significant degree, what is observed for the solution in beaker 3?

C. ...rate of vaporization increases
D. ...rate of vaporization decreases

Highlight for answer: D - now my question is why doesn't rate increase before decreasing

Please explain. I had this one backwards.

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[polyatomic]

2 Questions: 29 & 31

29. Above which beaker is the vapor pressure of methanol the greatest?

A. Beaker I
B. Beaker II
C. Beaker III
D. The vapor pressure of methanol is the same above all 3 beakers

Correct answer: B (highlight to show correct answer)

Passage Info.

All beakers are sealed.
Beaker 1 contains: 50g methanol and 50g ethanol
Beaker 2 contains: 50ml methanol and 50 ml ethanol
Beaker 3 contains: 1 mole methanol and 1 mole ethanol

Density of methanol: .7914g/ml (mw = 32; bp = 56 degrees C)
Density of ethanol: .7893g/ml (mw = 46; bp = 79 degrees C)

Total VP = XP(of methanol) + XP(of ethanol)
You'll have a greater amount of vapor pressure due to methanol whenever you have more moles of methanol compared to ethanol.

You need to compare to see which beaker has the highest ratio of moles of methanol to moles of ethanol. Try to do this with as few calculations as possible:

1 - I would think that in Beaker 1 I know that there are more moles of methanol than ethanol. We know this because whenever you have an equal mass of 2 compounds, the compound with the LOWER molecular weight has MORE moles. Remember this.

2 - I would eliminate Beaker 3 based upon that conclusion, because beaker 3 has equal moles of methanol and ethanol, thus Beaker 1 automatically has a higher methanol vapor pressure. Then I would eliminate choice D - because of what I just stated.

3 - Now I would standardize units between Beaker 1 and 2. I wouldn't convert both into moles because that requires extra steps in both calculations. Instead, I would just move everything into grams, because I easily know the relationship between grams and moles.

Beaker 2:
50 ml meth*0.7914 g/ml = mass of meth
50 ml eth* 0.7893 g/ml = mass of eth

Okay so those calculations would suck to figure out by hand. Thankfully...you don't have to! All you have to know here is that when you convert both into grams, you'll end up with MORE grams of methanol than grams of ethanol, (because 0.7914 > 0.7893).

4 - Now we can compare Beaker 1 to Beaker 2. Because in Beaker 2 we have a higher ratio of methanol to ethanol mass compared to Beaker 1, we know this equates to a higher ratio of moles of methanol to ethanol in Beaker 2, compared to Beaker 1. Therefore..............we have the a higher vapor pressure of methanol in Beaker 2.

 

[RiPSTONE]

For Q31, the answer explanation is pretty clear. As the methanol evaporates, there is more ethanol left in the solution, which has a lower VP. Therefore, the VP of the solution will decrease in comparison to when methanol was evaporating.