Stoich Questions

[FlipDoc2Be]

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Hey guys I have some stoich questions I could use some help with..

For a 2.0M aqueous salt solution with a density of 1.05 g/mL that is 9.17% solute by mass, what is the approximate molality?

A. 1.0m
B. 1.9m
C. 2.1m
D. 4.2m

The answer is C. Could someone show the math behind this? Thanks!

Which of the following will occur when the acidity of a saturated aqueous solution of Mg(OH)2 is increased?

A. Additional Mg(OH)2 will dissolve and the Ksp will increase.
B. Additional Mg(OH)2 will dissolve and the Ksp will be unchanged.
C. Some Mg(OH)2 will precipitate and the Ksp will decrease.
D. Some Mg(OH)2 will precipitate and the Ksp will be unchanged.

The answer B. I understand the Mg(OH)2 due to the increase in acidity but shouldn't the Ksp decrease?

Thanks again everyone.

 

[IntelInside]

Hey guys I have some stoich questions I could use some help with..

For a 2.0M aqueous salt solution with a density of 1.05 g/mL that is 9.17% solute by mass, what is the approximate molality?

A. 1.0m
B. 1.9m
C. 2.1m
D. 4.2m

The answer is C. Could someone show the math behind this? Thanks!

Which of the following will occur when the acidity of a saturated aqueous solution of Mg(OH)2 is increased?

A. Additional Mg(OH)2 will dissolve and the Ksp will increase.
B. Additional Mg(OH)2 will dissolve and the Ksp will be unchanged.
C. Some Mg(OH)2 will precipitate and the Ksp will decrease.
D. Some Mg(OH)2 will precipitate and the Ksp will be unchanged.

The answer B. I understand the Mg(OH)2 due to the increase in acidity but shouldn't the Ksp decrease?

Thanks again everyone.

For #1:

We assume we have 1 liter of the solution therefore we get 2mol of the salt. We know the density of this solution is 1.05g/ml or 1050 g/L we then have to calculate the portion of solute that is contributing to the weight. 1050 * 9.17 is approx 97. So then the mass of the pure solvent is around 950 grams. We then convert to kg (.95 kg) because the Molality formula is moles of solute/kg of solvent and solve.


For #2:

Ksp will remain unchanged, its an equilibrium constant and they depend only on temperature. The reason why the solubility of Mg(OH)2 will increase is because H+ ions will then associate with the hydroxide (OH-) ions and form water thus removing and allowing for an increase in ability to dissolve more Mg(OH)2

 

[BerkReviewTeach]

Hey guys I have some stoich questions I could use some help with..

For a 2.0M aqueous salt solution with a density of 1.05 g/mL that is 9.17% solute by mass, what is the approximate molality?

A. 1.0m
B. 1.9m
C. 2.1m
D. 4.2m

The answer is C. Could someone show the math behind this? Thanks!

Didn't the teacher go over this question in class? I ask, because it's a trick question meant to teach a critical strategy technique (and make a huge conceptual point in class). It's one of those questions that the teacher's outline emphasizes as something essential. There is not enough information in the question to answer it via math, so it's a chance to teach the students how to be pragmatic during their exam and take advantage of the multiple choice aspect.

The teacher was suppose to mention that the information you would need has to be fished forin the passage. Given that the question was part of the free-standing in class questions, the instructor was suppose to let the class struggle for thirty seconds and then interupt when no one was getting it. Showing the solution was meant to demonstrate that there was missing information (molecular mass of the salt) and occassionally you'd need to fish elsewhere. It was also suppose to point out that molality and molarity are usually close in value for aqueous solutions, because 1 L water is 1 kg. Because the solution density is slightly greater than 1, we can take a leap of faith that the amount of water in 1L of that solution is slightly less than a kg (implying that the salt occupies some space in the 1L of solution, so the solution contains slightly less than a full liter of water and thus slightly less than a full kg of water). The end result is that the molality is assumed to be slightly more than the molarity.

The major point the instructor was suppose to emphasize was that for the purposes of the MCAT, molality and molarity of aquesous solutions can be treated as essentially interchangeable, because the numerical values are so close.

Was the teacher rushed during that class?

 

[BerkReviewTeach]

Flip2be:

Please disregard my response above, because in a different thread you said you were not actually in the class and were just using the notes. Like I mentioned in that thread, you really shouldn't worry about the free standing questions in the packets, because many of them are like this one... without seeing the lecture firsthand, you are missing a key peice of information or point of the question. About half of the questions are designed to teach a test taking strategy or some conceptual point, so without seeing and hearing the follow up to the question, it might actually cause more harm than good.

Just stick to the passages in those handouts and don't worry about the single questions or diagrams. (Some of the diagrams are incomplete and students are suppose to write on them during lecture. Without the details, shortcuts, or tips, some won't make sense either.)

 

[Moosecakes]

Thanks for that post though berkreview!!

 

[IntelInside]

Didn't the teacher go over this question in class? I ask, because it's a trick question meant to teach a critical strategy technique (and make a huge conceptual point in class). It's one of those questions that the teacher's outline emphasizes as something essential. There is not enough information in the question to answer it via math, so it's a chance to teach the students how to be pragmatic during their exam and take advantage of the multiple choice aspect.

The teacher was suppose to mention that the information you would need has to be fished forin the passage. Given that the question was part of the free-standing in class questions, the instructor was suppose to let the class struggle for thirty seconds and then interupt when no one was getting it. Showing the solution was meant to demonstrate that there was missing information (molecular mass of the salt) and occassionally you'd need to fish elsewhere. It was also suppose to point out that molality and molarity are usually close in value for aqueous solutions, because 1 L water is 1 kg. Because the solution density is slightly greater than 1, we can take a leap of faith that the amount of water in 1L of that solution is slightly less than a kg (implying that the salt occupies some space in the 1L of solution, so the solution contains slightly less than a full liter of water and thus slightly less than a full kg of water). The end result is that the molality is assumed to be slightly more than the molarity.

The major point the instructor was suppose to emphasize was that for the purposes of the MCAT, molality and molarity of aquesous solutions can be treated as essentially interchangeable, because the numerical values are so close.

Was the teacher rushed during that class?

Ahh I solved it using math from my example above your post. How is there not enough data when I solved for it

 

[BerkReviewTeach]

Ahh I solved it using math from my example above your post. How is there not enough data when I solved for it

You know, you are absolutely correct. I didn't read the question as it was posted by the OP carefully enough and instead went by what was on the class handout back when I taught. The question didn't use to contain the %solute by mass, so it was not possible to solve the question. With the information now given, it is solved exactly as you showed.

I wonder why they added that into the question, except maybe to make a different point in terms of estimating molarity and molality.