March 28, 2024, 05:03:03 AM
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Topic: Determining molality of a solution given the density - assume no volume change?  (Read 7713 times)

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Offline riester

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I was asked to calculate the boiling point of a 1.5M aqueous solution of KI that has a density of 1.18g/mL. This hasn't been taught in class and I think I figured it out, but I assumed no change in volume upon the addition of KI when preparing the solution - is this acceptable?

I assumed the solution was 1.00g/mL water, meaning the remaining 0.18 g/mL would be KI. I then converted grams KI to moles KI (I got 1.08 x 10-3mol), then assumed no volume change when the solution was formed (meaning that the /mL of solution is the same as /mL of water). I used the density of water to convert the mL to kg, then calculated the molality of KI using this and the previously determined number of moles. Using the molality of KI (which I got to be 1.08m), the van't Hoff factor of KI, and the Kb value for water, I was able to determine 
:delta: tb and obtain a boiling point of about 101.1 degrees Celcius.

This should be correct from what we've learned so far, but nowhere in the problem does it say to assume no change in volume, and I'm not quite comfortable doing so. Does this method look correct?

Offline ramboacid

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There is a much more logical and simpler way to solve this using information already presented in the problem and with fewer assumptions.  The mistake you made was with calculating the molality. :)

The best way I found to solve these problems is to use the factor-label or dimensional-analysis methods. If you have your units correctly, it should work every time. 8)

You're given a 1.5M KI solution, so we can say you have 1.5 mol KI per 1000 ml of solution. Then, knowing the density of the solution, you can calculate the mass of the solution. At that point, you can calculate the number of moles of KI per 1000 g, or 1 kg, of solution and there it is you have the molality.

The calculation is setup as:

1.5 moles of KI          1 ml solution          1000 g solution
----------------  X  ---------------  X  -----------------
1000 ml solution        1.18 g solution         1 kg solution

Then we cancel units:

1.5 moles of KI          1 ml solution          1000 g solution
----------------  X  ---------------  X  -----------------
1000 ml solution        1.18 g solution         1 kg solution

Then we plug and chug to get 1.27 moles of KI per 1 kg solution, or 1.27 molal.

The trick is setting up the expression so that the units cancel. Using this method, all you have to do is let the unit cancellation guide you until you end up with the units you want.

I was actually struggling with this a few weeks back, and to get better I worked through these practice problems I found on the web:
http://www.chemteam.info/Solutions/Molality-from-density-and-percent.html
They didn't use the factor-label approach explicitly, but it turns out the same.

Using this method, you don't have to worry about a change in volume, because the density takes care of it for you. I believe the rest of you approach is correct.
"Opportunity is missed by most people because it is dressed in overalls and looks like work." - Thomas Edison

Offline riester

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Maybe I'm not understanding what you're getting at, but you were solving for molality as moles KI per kilogram of solution, rather than moles KI per kilogram of water (which I think it should have been). I didn't think this would work...but seeing your work gives me an idea.

Could you calculate the number of moles of KI in 1 kg of solution, convert that number of moles to grams of KI, then subtract that from the 1000 g of solution to get the grams (and then kilograms) of water in the solution, and then use the kilograms of water and moles of KI to get the molality as moles KI per kilogram of water? It could go like this:

1.5mol KI               1mL solution          1000g solution
---------------- X -------------- X --------------------  (canceling units)
1000 mL solution    1.18g solution         1kg solution

This would obtain 1.27 moles of KI per 1 kg solution. From this, we can calculate the grams of KI:

1.27mol KI       166.00g KI
------------ X ------------  (canceling units)
1kg solution      1mol KI

This would obtain 210.82 g of KI per kg of solution - this would mean that the other .78918 kg of solution would be water, right? We could then use the number of moles and the kg of water to obtain the ratio of moles solute to kilograms solvent:

1.27mol KI
---------------- = 1.61 molal
.78918 kg water

This molality could then be used in the boiling point calculations. This makes much more sense than what I was doing beforehand. Does it look good to you?

Offline ramboacid

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Quote
Maybe I'm not understanding what you're getting at, but you were solving for molality as moles KI per kilogram of solution, rather than moles KI per kilogram of water (which I think it should have been).

 :o *facepalm*
Yikes. My bad entirely. Good thing you checked that. I checked my notes and saw that I've forgotten some of my stuff. Looks like I still need to review a bit for the AP test this monday ::)

Your reasoning is perfect. An easy way to do this mathematically would be to calculate the weight of the solution using the density and then subtract the number of moles afterward. We would assume 1000 ml of solution initially to make the math easier.

                            1.18 g solution
1000 ml solution  X  ---------------  =  1180 g solution
                             1 ml solution

We're told it is a 1.5 M solution, so there are 1.5 moles of KI in 1000 ml of solution. The weight of the KI is
                     166 g KI
1.5 mol KI  X  ----------- = 249 g KI
                     1 mol KI

Subtracting gives the total weight of water:

1180 g solution - 249 g solute = 931 g solvent

Then all that's left is to simply divide to find the molality:

  1.5 mol KI
-------------- = 1.61 molal
0.931 kg water

I personally think this method, although exactly equivalent to yours, is less tedious because you don't have to specifically recalculate the data that's given to you for 1 kg water until the end when its not an explicit step. But as always, just do whatever floats your boat  ;D

I just want to say I am really sorry if I had you confused. I should've checked my math first before trying to help you.  Probably shouldn't have been expounding on the virtues of the factor-label method and instead rereading my post  :P
"Opportunity is missed by most people because it is dressed in overalls and looks like work." - Thomas Edison

Offline riester

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Hey, it's fine - I've accidentally used moles solute per kilogram solution instead of moles solute per kilogram solvent way more times than I'd like to admit, and I was happy to be able to work this problem out with someone. Also, I do like your method better - it's faster.

Good luck with the AP test! :) I'll be taking the chem AP test next year, but I still have to study for AP comp sci this year. :(

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