Vapor Pressure Question

[arc5005]

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PASSAGE <-- Passage

TL;DR:
Beaker I: 50 mg ethanol, 50 mg methanol
Beaker II: 50 ml ethanol, 50 ml methanol
Beaker III. 1 mole ethanol, 1 mole methanol

If the vapor pressure of pure methanol is greater than the vapor pressure of pure ethanol, then what must be true about the relative temperatures of each solution in order to have equal total vapor pressures above all three beakers?


A. TI > TII > TIII
B. TIII > TI > TII
C. TI > TIII > TII
D. TII > TI > TIII



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B) TIII > TI > TII

The solution richest in methanol has the greatest vapor pressure. The greatest mole fraction of methanol is found in Beaker II, so at equal temperatures, the vapor pressure above Beaker II is the greatest. The mole fraction of methanol is greater in Beaker I than in Beaker III, so at equal temperatures, the vapor pressure is given by:

Pvapor Beaker II > Pvapor Beaker I > Pvapor Beaker III

To increase the vapore pressure above the solution, the temperature must be increased, so for the Beaker III solution to exert the same vapor pressure as the Beaker II solution, the temperature of the solution in Beaker III must be increased. This means that the temperature of the solution in Beaker III must be the greatest, if the vapor pressures above all three beakers were equal.

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What exactly is the process to think about for Beaker II?

 

[Lawpy]

The passage mentions Raoult's Law, which states that the vapor pressure of a component in a mixture = mole fraction of that component in mixture * pure vapor pressure of that component. For a methanol-ethanol mixture, Raoult's Law shows that:

vapor pressure of methanol in mixture = mole fraction of methanol * vapor pressure of pure methanol
vapor pressure of ethanol in mixture = mole fraction of ethanol * vapor pressure of pure ethanol

The question says vapor pressure of pure methanol > vapor pressure of pure ethanol. This means that the beaker with the highest amount of methanol will have the highest vapor pressure, so we can focus on the mole fraction of methanol.

The passage defines the mole fraction = moles of component / moles of solution. For the methanol-ethanol mixture, this means:

mole fraction of methanol = moles of methanol / (moles of methanol + moles of ethanol)

For Beaker I (use the molecular weight to convert from mass to moles):

50 mg ethanol * (1 g ethanol) / (1000 mg ethanol) * (1 mol ethanol) / (46.07 g ethanol) = (1/20)*(1/46.07) mol ethanol
50 mg methanol * (1 g methanol) / (1000 mg methanol) * (1 mol methanol) / (32.04 g methanol) = (1/20)*(1/32.04) mol methanol

mole fraction of methanol = ((1/20)*(1/32.04))/((1/20)*(1/32.04)+(1/20)*(1/46.07)) = (1/32.04) / (1/32.04 + 1/46.07) -->
mole fraction of methanol = (1/32.04) / ((46.07+32.04)/(32.04*46.07)) -->
mole fraction of methanol = (1/32.04) * (32.04*46.07)/(46.07+32.04) -->
mole fraction of methanol = 46.07/(46.07+32.04) -->
mole fraction of methanol = 46.07/78.11

For Beaker II (use the density to convert from volume to mass and molecular weight to convert from mass to moles):

50 mL ethanol * (0.7893 g ethanol) / (1 mL ethanol) * (1 mol ethanol) / (46.07 g ethanol) = (50 * 0.7893) / 46.07 mol ethanol
50 mL methanol * (0.7914 g methanol) / (1 mL methanol) * (1 mol methanol) / (32.04 g methanol) = (50 * 0.7914) / 32.04 mol methanol

mole fraction of methanol = ((50*0.7914)/32.04)/((50*0.7914)/32.04+(50*0.7893)/46.07) = (0.7914/32.04)/(0.7914/32.04+0.7893/46.07) -->
mole fraction of methanol = (0.7914/32.04)/((0.7914*46.07+0.7893*32.04)/(32.04*46.07)) -->
mole fraction of methanol = (0.7914/32.04) * (32.04*46.07) / (0.7914*46.07+0.7893*32.04) -->
mole fraction of methanol = (0.7914*46.07)/(0.7914*46.07+0.7893*32.04) -->
mole fraction of methanol = 46.07/(46.07+(0.7893/0.7914)*32.04) (divide both the numerator and denominator by 0.7914).

For Beaker III:

mole fraction of methanol = 1/(1+1) = 1/2

Raoult's Law says the beaker with the highest mole fraction of methanol has the highest vapor pressure. We can immediately see that Beaker I has a higher vapor pressure than Beaker III because 46.07/78.11 > 1/2 (46.07/78.11 is similar to 46/78, and 46/78 > 46/80 > 40/80 > 1/2).

Beaker II has a higher vapor pressure than Beaker I because:

0.7893/0.7914 < 1 -->
(0.7893/0.7914)*32.04 < 32.04 -->
46.07+(0.7893/0.7914)*32.04 < 46.07 + 32.04 -->
46.07+(0.7893/0.7914)*32.04 < 78.11 -->
1/(46.07+(0.7893/0.7914)*32.04) > 1/78.11 -->
46.07/(46.07+(0.7893/0.7914)*32.04) > 46.07/78.11

Sorry for the complicated calculations: it's definitely easier to simply use a calculator, but it's good practice to work these calculations out by hand since the MCAT doesn't allow calculators. It's good practice, even though you will see far easier numbers to work with on test day.

Ranking by vapor pressures, we see Beaker II > Beaker I > Beaker III. But if the vapor pressures above each of these beakers have to be equal, the relative temperatures have to change to counteract the differences in changes in vapor pressure seen due to different mole fractions. In other words, because Beaker II has the highest mole fraction of methanol, it has to have the lowest relative temperature to have the same total vapor pressure. Likewise, Beaker III has the lowest mole fraction of methanol so it has to have the highest relative temperature.

Ranking by relative temperatures, we get Beaker III > Beaker I > Beaker II.

 

[arc5005]

The passage mentions Raoult's Law, which states that the vapor pressure of a component in a mixture = mole fraction of that component in mixture * pure vapor pressure of that component. For a methanol-ethanol mixture, Raoult's Law shows that:

vapor pressure of methanol in mixture = mole fraction of methanol * vapor pressure of pure methanol
vapor pressure of ethanol in mixture = mole fraction of ethanol * vapor pressure of pure ethanol

The question says vapor pressure of pure methanol > vapor pressure of pure ethanol. This means that the beaker with the highest amount of methanol will have the highest vapor pressure, so we can focus on the mole fraction of methanol.

The passage defines the mole fraction = moles of component / moles of solution. For the methanol-ethanol mixture, this means:

mole fraction of methanol = moles of methanol / (moles of methanol + moles of ethanol)

For Beaker I (use the molecular weight to convert from mass to moles):

50 mg ethanol * (1 g ethanol) / (1000 mg ethanol) * (1 mol ethanol) / (46.07 g ethanol) = (1/20)*(1/46.07) mol ethanol
50 mg methanol * (1 g methanol) / (1000 mg methanol) * (1 mol methanol) / (32.04 g methanol) = (1/20)*(1/32.04) mol methanol

mole fraction of methanol = ((1/20)*(1/32.04))/((1/20)*(1/32.04)+(1/20)*(1/46.07)) = (1/32.04) / (1/32.04 + 1/46.07) -->
mole fraction of methanol = (1/32.04) / ((46.07+32.04)/(32.04*46.07)) -->
mole fraction of methanol = (1/32.04) * (32.04*46.07)/(46.07+32.04) -->
mole fraction of methanol = 46.07/(46.07+32.04) -->
mole fraction of methanol = 46.07/78.11

For Beaker II (use the density to convert from volume to mass and molecular weight to convert from mass to moles):

50 mL ethanol * (0.7893 g ethanol) / (1 mL ethanol) * (1 mol ethanol) / (46.07 g ethanol) = (50 * 0.7893) / 46.07 mol ethanol
50 mL methanol * (0.7914 g methanol) / (1 mL methanol) * (1 mol methanol) / (32.04 g methanol) = (50 * 0.7914) / 32.04 mol methanol

mole fraction of methanol = ((50*0.7914)/32.04)/((50*0.7914)/32.04+(50*0.7893)/46.07) = (0.7914/32.04)/(0.7914/32.04+0.7893/46.07) -->
mole fraction of methanol = (0.7914/32.04)/((0.7914*46.07+0.7893*32.04)/(32.04*46.07)) -->
mole fraction of methanol = (0.7914/32.04) * (32.04*46.07) / (0.7914*46.07+0.7893*32.04) -->
mole fraction of methanol = (0.7914*46.07)/(0.7914*46.07+0.7893*32.04) -->
mole fraction of methanol = 46.07/(46.07+(0.7893/0.7914)*32.04) (divide both the numerator and denominator by 0.7914).

For Beaker III:

mole fraction of methanol = 1/(1+1) = 1/2

Raoult's Law says the beaker with the highest mole fraction of methanol has the highest vapor pressure. We can immediately see that Beaker I has a higher vapor pressure than Beaker III because 46.07/78.11 > 1/2 (46.07/78.11 is similar to 46/78, and 46/78 > 46/80 > 40/80 > 1/2).

Beaker II has a higher vapor pressure than Beaker I because:

0.7893/0.7914 < 1 -->
(0.7893/0.7914)*32.04 < 32.04 -->
46.07+(0.7893/0.7914)*32.04 < 46.07 + 32.04 -->
46.07+(0.7893/0.7914)*32.04 < 78.11 -->
1/(46.07+(0.7893/0.7914)*32.04) > 1/78.11 -->
46.07/(46.07+(0.7893/0.7914)*32.04) > 46.07/78.11

Sorry for the complicated calculations: it's definitely easier to simply use a calculator, but it's good practice to work these calculations out by hand since the MCAT doesn't allow calculators. It's good practice, even though you will see far easier numbers to work with on test day.

Ranking by vapor pressures, we see Beaker II > Beaker I > Beaker III. But if the vapor pressures above each of these beakers have to be equal, the relative temperatures have to change to counteract the differences in changes in vapor pressure seen due to different mole fractions. In other words, because Beaker II has the highest mole fraction of methanol, it has to have the lowest relative temperature to have the same total vapor pressure. Likewise, Beaker III has the lowest mole fraction of methanol so it has to have the highest relative temperature.

Ranking by relative temperatures, we get Beaker III > Beaker I > Beaker II.

Thank you for doing the actual math, it helps to fully understand it and actually see the numbers!! I think I've finally gotten this problem down to understanding it without having to do all the complicated calculations, since there won't be time for that on the MCAT.