TBR CBT 5 Physical Sciences

[ModernAlchemist]

I'm not trying to be too presumptuous, but I'm going to guess that this was just a hard section and not worry too much about the score I got. I made this thread because I had many questions on this section and thought other people could post here too if they needed help.

Regarding the first passage:

Q3 and Q4 confused me. I don't really know how they arrived at their answers even after trying to work through the explanations. For those of you who don't feel like looking it up...

Q3: How many grams of glucose must be added to 0.5 kg of water to raise the boiling point by 0.10 ˚C, given that the value of i is 1 and the value of γ for glucose is roughly equal to that of formic acid? Choices were 5, 15, 90 and 180 grams.
*How do you figure this out? It says that formic acid changed the boiling point 0.6 ˚C and that from that, you can somehow deduce the answer.

Q4: The vapor pressure of water under standard conditions is 23.8 torr. What is the vapor pressure under standard conditions of a solution made by adding 10 grams of NaCl to 100 grams of water? Given VP of NaCl at 1.0 m is 21.1 torr.
*I don't understand how the VP of the soln is 20.3 torr.

More specifically, I want to have a definite strategy when pursuing VP and BP questions on the MCAT and not just be confused.

Can anyone help? THANK YOU

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[ChocoMoo]

I'm glad I'm not the only one who thought this PS section was difficult. I had a hard time with these two questions too and marked both of them.

I still don't understand how to solve Q3, but I can try to help out for Q4.

The way I think of vapor pressure is by imagining water in a container or something. If you were to zoom up super close to the air/liquid boundary, there would be some molecules of water that happened to gain enough kinetic energy to 'break free' from the liquid and escape as gas. These molecules make up the vapor pressure of the liquid. The more molecules that break free, the greater the VP. When you add something like salt to water, the salt molecules get in the way of the water molecules that might otherwise have enough kinetic energy to escape the liquid. Therefore your vapor pressure decreases (less water molecules can escape as gas). Knowing that adding salt to water decreases the VP, you can automatically eliminate A (26.4 torr) and choice B (23.8 torr) because they're bigger values than the standard VP of pure water that's given.

From the table they gave, you know that 1m NaCl has a VP of 21.1 torr. The units for m are moles solute/kg solvent. So 1m = 1 mol NaCl / 1 kg water.

They tell you that you have 10 grams of NaCl and 100 grams of water. Using some stoichiometry you can find that this is the same as 100 g of NaCl for 1 kg of water. 100 g of NaCl is equal to almost 2 moles of NaCl, so your molal for your solution is going to be something like ~2m (~2 moles NaCl/1 kg water). This number is greater than what was given in the table (1m), which means your solution has more moles of salt. We know salt decreases VP, so the VP of our solution must be less than the VP for a 1m solution. The VP of the 1m of solution was given as 21.1 torr, so our answer must be less than that. Answer choice D is the only one that fits.

Hope that made sense! And sry for the length.

 

[BerkReviewTeach]

Q3: How many grams of glucose must be added to 0.5 kg of water to raise the boiling point by 0.10 ?C, given that the value of i is 1 and the value of ? for glucose is roughly equal to that of formic acid? Choices were 5, 15, 90 and 180 grams.
*How do you figure this out? It says that formic acid changed the boiling point 0.6 ?C and that from that, you can somehow deduce the answer.

The correlation between formic acid and glucose is that they both have i=1 and the same gamma value, so equal molal solutions of formic acid and glucose should have the same boiling points. They want you to deduce the target molality for the glucose solution from the formic acid information. A 1.0 molal formic acid solution has a bp of 100.6 degreesC, so 1.0 molal corresponds with a deltaT of 0.6 degreesC. To get a deltaT of 0.10 degreesC, you would only need one-sixth molal formic acid. This means that we need 0.167 molal glucose in order to have a boiling point elevation (deltaT) of 0.1 degreesC.

The question tells us that there is 0.5 kg water, and molality is the moles solute over kg solvent. So 1/6 = moles glucose/0.5. Multiplying both sides by 0.5 tells us that we need 1/12 moles glucose in 0.5 kg water to get a 1/6 molal solution, which will have a deltaT of 0.1 degreesC.

Glucose has a MW of 180, and 1/12 x 180 = 15 grams.

 

[BerkReviewTeach]

I'm glad I'm not the only one who thought this PS section was difficult. I had a hard time with these two questions too and marked both of them.

I still don't understand how to solve Q3, but I can try to help out for Q4.

The way I think of vapor pressure is by imagining water in a container or something. If you were to zoom up super close to the air/liquid boundary, there would be some molecules of water that happened to gain enough kinetic energy to 'break free' from the liquid and escape as gas. These molecules make up the vapor pressure of the liquid. The more molecules that break free, the greater the VP. When you add something like salt to water, the salt molecules get in the way of the water molecules that might otherwise have enough kinetic energy to escape the liquid. Therefore your vapor pressure decreases (less water molecules can escape as gas). Knowing that adding salt to water decreases the VP, you can automatically eliminate A (26.4 torr) and choice B (23.8 torr) because they're bigger values than the standard VP of pure water that's given.

From the table they gave, you know that 1m NaCl has a VP of 21.1 torr. The units for m are moles solute/kg solvent. So 1m = 1 mol NaCl / 1 kg water.

They tell you that you have 10 grams of NaCl and 100 grams of water. Using some stoichiometry you can find that this is the same as 100 g of NaCl for 1 kg of water. 100 g of NaCl is equal to almost 2 moles of NaCl, so your molal for your solution is going to be something like ~2m (~2 moles NaCl/1 kg water). This number is greater than what was given in the table (1m), which means your solution has more moles of salt. We know salt decreases VP, so the VP of our solution must be less than the VP for a 1m solution. The VP of the 1m of solution was given as 21.1 torr, so our answer must be less than that. Answer choice D is the only one that fits.

Hope that made sense! And sry for the length.

Right on ChocoMoo, that is an excellent explanation!

 

[jchen08]

Would you mind explaining how you deduced that "To get a deltaT of 0.10 degreesC, you would only need one-sixth molal formic acid" ? ... I'm a visual person so reading this without seeing equation manipulation is hard to visualize. thanks for the help!

 

[BerkReviewTeach]

Would you mind explaining how you deduced that "To get a deltaT of 0.10 degreesC, you would only need one-sixth molal formic acid" ? ... I'm a visual person so reading this without seeing equation manipulation is hard to visualize. thanks for the help!

The table says that a 1.0 molal solution has a bp 100.6 degreesC, so we know that 1.0 m formic acid results in a deltaT of 0.6 degreesC. We want a deltaT of 0.1 degrees, which is one-sixth of 0.6. Molality is proportional to deltaT, so one-sixth the deltaT equates to one-sixth the molal, which happens to be 1/6 molal in this case.

 

[brood910]

Bump. This is a terrible question. Just for anyone who took TBT #5, AAMC will never ask a question like this. How the hell are we even supposed to know the MW of glucose?

lol.. so many errors in explanations in this FL as well.
TBR maybe hires the laziest ppl.

 

[sky1025]

Bump. This is a terrible question. Just for anyone who took TBT #5, AAMC will never ask a question like this. How the hell are we even supposed to know the MW of glucose?

lol.. so many errors in explanations in this FL as well.
TBR maybe hires the laziest ppl.

What do you mean? I agree with the part that AAMC is unlikely to ask such a question, but anyone who's studying for the MCAT should know what the MW of glucose is. That's a simple calculation from its molecular formula, and there is a periodic table given to you even if you didn't know the molecular masses of C, H, O.