Study Guide Question....molarity

[Fuzzychickens]

I have a study guide that asks how many grams of a compound are needed to make 90 grams of a 0.3 molar solution. The molecular weight of the compound times the mole ratio is 102.6g/(L of solution). The way I look at it is you mulitply the ratio of 90 grams to 1000 grams by 102.6g/L

90/1000 x 102.6g/L

The book says the answer is to add 102.6g/L to the liter (1000grams + 102.6g).

90g/(1000g + 102.6g) x 102.6g/L <----but how the heck is that any more accurate?, it seems to be a really fudgy solution considering it's not really mole per liter anymore but mole per liter plus how ever many liters 102.6 grams of the compound takes up.

I know neither of these methods are absolutely mathmatically correct, but I feel the book is using the least accurate method. Isn't it better to make the assumption that the compound will roughly add a millimeter per gram when dissolved into solution and calculate it as part of the 1000g of solution rather than adding it to the 1000g (liter) and effectively giving you a result less dilute than true .3 molar solution?

Which quick and dirty method would the PCAT subscribe to?

---

Members don't see this ad.

 

[Farmercyst]

I have a study guide that asks how many grams of a compound are needed to make 90 grams of a 0.3 molar solution. The molecular weight of the compound times the mole ratio is 102.6g/(L of solvent).

It doesn't make sense to me as worded.

 

[omnione]

This question is stupid....but I'll save my explanation for later. Here's how I interpreted the question and please correct me if I'm wrong on something:

1. You are in a chem lab. On one side you have solute in powder/solid form. On the other side of you you have your solvent, presumably water given the answer you provided.

2. You want to make a 0.3 M solution that weighs 90g (Why? But I'll play along......)

3. You are told that MW X 0.3 mol/L = 102.6 g/L(this part was confusing, poorly worded, and incorrect given that Molarity is moles per liter of solution not solvent). In principle, 1 L of 0.3M solution resulted from 102.6 g of your compound dissolved in water.

But if you make a liter of this stuff, that's going to weigh more than the 90g that you want. Specifically, it would weigh 102.6 g + 1000g (the mass of water at 25C) = 1102.6g.

So, I guess the idea is to downsize the amount of solution you make so that:

A) The molarity is still 0.3 M while,
B) The liquid (making an assumption once again that it turns out to be liquid) weighs 90g

Thus, the book suggests a downsize factor (not dilution factor since we aren't trying to change the molarity of the solution) logically would be:

the mass you want/the mass you are given or know

90g/ (1000g + 102.6g).

Here's where the book messes up a little from my interpretation. You would take that downsize factor and you multiply it by the mass of solute powder used in making a liter of 0.3M solution to get the appropriate mass of solute to use. You would also multiply that factor to the mass of total solution (I corrected the book's mistake here) used in making a liter of 0.3M solution too because you have to downsize that by an equal proportion to keep the solution at 0.3M. Though the book comes up with the correct numerical value, the units imply that the you are changing the concentration. The way the book should have worded the answer to make sense is:

90 g/(1000g + 102.6 g) * 102.6 g = The same number that the book gives you.

On a side note, you would use 90 g/(1000g + 102.6 g) * (1000g of water - answer above) to get the mass of water to use as solvent.

Of course, you would use the correct amount of solvent to get up to 0.3 M and 90 g of solution mass. But the question assumes that you will so the the book just wants to know how much of that solute powder you should use. That answer will tell you how much of that solute powder to use if you want to make 90 g of a 0.3M solution assuming that you are using water as a solvent and the density of water is 1.00 g/ml.

Tell me if this is what the book was trying to test here since this question was poorly worded and impractical for a PCAT question let alone in a real-life chem lab. If I'm right, then I'll go into a bash-fest about the problem.

 

[Bhavesh]

1. You are in a chem lab. On one side you have solute in powder/solid form. On the other side of you you have your solvent, presumably water given the answer you provided.

2. You want to make a 0.3 M solution that weighs 90g (Why? But I'll play along......)

3. You are told that MW X 0.3 mol/L = 102.6 g/L(this part was confusing, poorly worded, and incorrect given that Molarity is moles per liter of solution not solvent). In principle, 1 L of 0.3M solution resulted from 102.6 g of your compound dissolved in water.

But if you make a liter of this stuff, that's going to weigh more than the 90g that you want. Specifically, it would weigh 102.6 g + 1000g (the mass of water at 25C) = 1102.6g.

So, I guess the idea is to downsize the amount of solution you make so that:

A) The molarity is still 0.3 M while,
B) The liquid (making an assumption once again that it turns out to be liquid) weighs 90g

Thus, the book suggests a downsize factor (not dilution factor since we aren't trying to change the molarity of the solution) logically would be:

the mass you want/the mass you are given or know

90g/ (1000g + 102.6g).

Here's where the book messes up a little from my interpretation. You would take that downsize factor and you multiply it by the mass of solute powder used in making a liter of 0.3M solution to get the appropriate mass of solute to use. You would also multiply that factor to the mass of total solution (I corrected the book's mistake here) used in making a liter of 0.3M solution too because you have to downsize that by an equal proportion to keep the solution at 0.3M. Though the book comes up with the correct numerical value, the units imply that the you are changing the concentration. The way the book should have worded the answer to make sense is:

90 g/(1000g + 102.6 g) * 102.6 g = The same number that the book gives you.

On a side note, you would use 90 g/(1000g + 102.6 g) * (1000g of water - answer above) to get the mass of water to use as solvent.

this does make sense, and I think the problem ends right there

 

[binghamkid]

*boggle* why bother setting something that complicated up?

Just boil this problem down to a units problem.

Sinc they want 90 grams of 0.3 molarity, write it out.

90 g * 0.3 mol/1L = 102.6 g/1L * N g.

Solve for N to find # of grams.

A little bit of mathematical canceling and you've got your answer.

the left side can be simplified by moving the decimal point:

9g*3mol/1L = 102.6g/1L * N g

If you look on the right, the 102.6 is a multiple of 3, so cancel out the 3's ( it's also a multiple of 9, but if you're aren't sure, go with the 3)

9g*mol/1L = 34.2g/1L * N g

And you can do it again ( the multiple of 9 can eliminate these two steps into one)

3 g*mol/1L = 11.4 g/1L * N g

Cancel out all the units.

N = 3/(11.4)

If you simplify the problem this way, it's much easier to pick out your answer. Without calculating I can already tell the answer is around 0.25 ( a little less than that.) The PCAT will never give you answers that are very close to each other unless there's a reason to (like you need to account for sig figs). Knowing my estimated answer I can probably narrow the answer down to two, maybe even one.

Don't get caught up finding the exact answer for the test. You don't need to show your work for the exam. Think back to your middle school days when you tried to take shortcuts on your tests and you'll do better.

 

[omnione]

*boggle* why bother setting something that complicated up?

Well, I never said that I would set up the problem that way on the real PCAT of course.

When I explain problems on this board, I dissect the problem just to make sure the OP gets the whole picture. The problem with dumbing down the problem and/or just listing the shorthand, shortcutted method to solve the problem is that it assumes that the person will still understand the science going on when you just add or multiply numbers. Despite the frenetic pace of the PCAT, it's better to understand the science and use that to solve problems as opposed to memorizing "quick-fix" solutions in a rote manner and hope that you used the correct method in a problem.

As I said before, of course you want to shorthand this stuff on the real exam. But since this is a discussion board for people asking for help, help should be more than quick simplifications.

 

[binghamkid]

Well, I never said that I would set up the problem that way on the real PCAT of course.

When I explain problems on this board, I dissect the problem just to make sure the OP gets the whole picture. The problem with dumbing down the problem and/or just listing the shorthand, shortcutted method to solve the problem is that it assumes that the person will still understand the science going on when you just add or multiply numbers. Despite the frenetic pace of the PCAT, it's better to understand the science and use that to solve problems as opposed to memorizing "quick-fix" solutions in a rote manner and hope that you used the correct method in a problem.

As I said before, of course you want to shorthand this stuff on the real exam. But since this is a discussion board for people asking for help, help should be more than quick simplifications.

Lol i was responding to the book's complicated way of doing things. It's a good thing to explain the science concepts behind it, but since you did it already =D I didn't see fit to do it again

 

[omnione]

Lol i was responding to the book's complicated way of doing things. It's a good thing to explain the science concepts behind it, but since you did it already =D I didn't see fit to do it again

To be honest, I think this study guide is full of crap. Though the PCAT often asks trivia-like or detail questions, the science problems do reflect stuff I've seen in science courses and labs. I don't think I've seen a Harcourt PCAT question like this, but I could be wrong.

In an undergraduate lab setting, or maybe even a research lab, I can't think of a practical reason why anyone would want to make a certain mass of liquid solution. Generally, we are more concerned with the molarity of the solution. Mass is more common for solid products from a series of chemical reactions. If I do want to make a certain amount of a given molar of solution, I simply establish the amount of solution I want and use my desired molarity and dimensional analysis to calculate the mass of solid solute I need.

More likely, the PCAT will ask that type of molarity question where you have to calculate the amount of solute used in a x M solution with a complication here or there. The only reason this study guide uses this obscure problem seems to be that it's just a units problem or it's making a obscure problem for the sake of practicing weird problems. From my real PCAT experience, all of the chemistry questions covered topics and methods found in my chemistry courses I took.

 

[Fuzzychickens]

3. You are told that MW X 0.3 mol/L = 102.6 g/L(this part was confusing, poorly worded, and incorrect given that Molarity is moles per liter of solution not solvent). In principle, 1 L of 0.3M solution resulted from 102.6 g of your compound dissolved in water.

Sorry, I meant solution. I edited that.

The question the book asks "how many grams of a compound
(mw = 342g/mole) are needed to make 90 grams of a 0.3 molar solution"
The solvent is water.

The final answer the book gives: 90 g/(1000g + 102.6 g) * 102.6 g

My beef is this method results in a solution that is slightly less than .3 m. Their answer gives you 8.37g of solute that will be part of a 90g solution.
That comes to (8.37g/((342g/mole)(.090L)) = .2719 molarity.

If I had approached their problem my way I'd get, 90g/1000g * 102.6
= 9.234g solute.

Giving me 9.234g/((342g/mole)(.090L)) = .3 molarity

Of course, I make the assumption that the solute when added will have roughly the same density of water (1g/ml) - take up the same amount of volume per weight.

Anyways, I've seen textbooks do these types of problem both ways. Kinda bugs me. The whole question is silly (as you said) since no one would ever approach it in this way, but I find their solution to be incorrect.

 

[omnione]

The question the book asks "how many grams of a compound
(mw = 342g/mole) are needed to make 90 grams of a 0.3 molar solution"
The solvent is water.

The final answer the book gives: 90 g/(1000g + 102.6 g) * 102.6 g

That's much better wording....

I figured out how they did the problem as I would just use a simple proportion given that information:

x/90 g total = (0.3 * 342) g compound/ (1000 + 0.3 * 342)g total

x/90g total = 102.6 g compound/ 1102.6 g total

x = 90 g total *(102.6 g compound/1102.6 g total)

x = the amount of compound you want to use.

That process seems consistent with your book's answer. Basically, they used a proportion through a ratio of mass of compound/total mass between 1 L of 0.3 M solution and 90g of desired solution.

 

[Fuzzychickens]

That's much better wording....

I figured out how they did the problem as I would just use a simple proportion given that information:

x/90 g total = (0.3 * 342) g compound/ (1000 + 0.3 * 342)g total

x/90g total = 102.6 g compound/ 1102.6 g total

x = 90 g total *(102.6 g compound/1102.6 g total)

x = the amount of compound you want to use.

That process seems consistent with your book's answer. Basically, they used a proportion through a ratio of mass of compound/total mass between 1 L of 0.3 M solution and 90g of desired solution.

Yes, but I still think the book is wrong. If the book is right then the molarity of the 90g of solution (8.37g of which is the solute) should be 0.3.

But it's not. It comes to: (8.37g/((342g/mole)(.090L)) = 0.2719 molarity

If I do it the other way, assuming the solute is the same density as water, I get: 90g/1000g * 102.6 = 9.234g solute.

Giving me a molarity of .3 (correct) : 9.234g/((342g/mole)(.090L)) = .3 molarity

Anyway, this question was from a Kaplans 2007 guide...slightly different value used, but the question is the same. I just think they have it wrong. If they say it's supposed to be a 0.3 molar solution, then the math should support that and it doesn't.

 

[omnione]

Yes, but I still think the book is wrong. If the book is right then the molarity of the 90g of solution (8.37g of which is the solute) should be 0.3.

Anyway, this question was from a Kaplans 2007 guide...slightly different value used, but the question is the same. I just think they have it wrong. If they say it's supposed to be a 0.3 molar solution, then the math should support that and it doesn't.

\

I know that, I was just trying to figure out how they did the problem

 

[Fuzzychickens]

Ah ok. I was just afraid the book was right and I was having a never ending brain fart.