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### Topic: Conversion (molarity, molality, %, etc)  (Read 12438 times)

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#### WhoCares357

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##### Conversion (molarity, molality, %, etc)
« on: August 28, 2010, 09:06:43 PM »
I have this problem for homework:
Quote
How many mL of 1M acetic acid (C2H4O2) and .2 molal acetic acid need to be added to make 100 mL of vinegar (5% acetic acid)? Assume 1g/mL density of all solutions.

I started by listing everything that is given: density = 1g/mL; 1M solute = 1 mol solute/1 L solution; .2 molal solute = .2mol solute/1kg solvent (H2O); 60.052 g/mol solute; 18.000 g/mol solvent; .1L total solution. I am looking for the mL of solute in .1L of solution.

My first step was to use the Molarity to figure out the number of moles of the solute: 1M solute = .1 mol solute/.1L solution.
Then I used the number of moles to find the number of grams in the solution: .1mol x 60.052g/mol = 6.0052g solute.
Finally I converted the mass to volume using density: 6.0052g/(1g/mL) = 6.0052mL.

However I am pretty sure that this answer is wrong. When I insert the .1 mol of solute into the molality formula I get a totally different answer.
.2 molal = .1mol/500g -> 500gx.05/.95 = 26.315g -> 26.315g/(1g/mL) = 26.315mL.

26.315mL=/= 6.0052mL

Can anyone show me where I went wrong. It's possible that I may have just overthought it, but the answer just doesn't seem right to me.

Thank you.

#### opti384

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #1 on: August 28, 2010, 09:40:27 PM »
Are you sure that 100ml of vinegar (5% acetic acid) is 1M? I think by saying "My first step was to use the Molarity to figure out the number of moles of the solute: 1M solute = .1 mol solute/.1L solution," you are starting from assuming that 100ml of 5% acetic acid is 1M.

#### WhoCares357

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #2 on: August 28, 2010, 10:16:40 PM »
I just assumed that molarity was a constant. And since I was given the volume of the solution, I figured the only variable was the number of moles of solute. What does the 1M actually mean?

Edit: Am I at least right in assuming that the 1M applies to the solute and not the solvent/solution?
Edit 2: Also molarity = moles of solute / liters of solution. This way I didn't assume that .1L applied to the acetic acid but rather to the solution.

#### opti384

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #3 on: August 28, 2010, 10:18:29 PM »
Molarity(M) is a concentration expressed in 1mol/1L= 1M

#### WhoCares357

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #4 on: August 28, 2010, 10:28:35 PM »
Molarity(M) is a concentration expressed in 1mol/1L= 1M

Yes, but also 1M = 1 mol of solute / 1 L of solution

Since 1M is given as a constant (for the solute) and .1L is given as a constant (for the solution [vinegar]), I figured the only variable in the equation was the number of moles of the solute (acetic acid).

So 1M (acetic acid) = 1 mol of acetic acid / 1 Liter of vinegar

#### opti384

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #5 on: August 28, 2010, 10:35:57 PM »
Molarity(M) is a concentration expressed in 1mol/1L= 1M

Yes, but also 1M = 1 mol of solute / 1 L of solution

Since 1M is given as a constant (for the solute) and .1L is given as a constant (for the solution [vinegar]), I figured the only variable in the equation was the number of moles of the solute (acetic acid).

So 1M (acetic acid) = 1 mol of acetic acid / 1 Liter of vinegar

What I was trying to say was with 100ml of vinegar we can't assume it's 0.1 mol of acetic acid. In order to solve this problem you have to figure out how much mols of acetic acid is included n 100ml of vinegar (5% acetic acid). There's no indication that 100ml of vinegar is 1M.

#### WhoCares357

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #6 on: August 28, 2010, 10:46:11 PM »
So I disregard the molarity?

I tried it again, this time converting mL of solution to grams of solution (100). Then I plugged in the mass into the molality equation. I got .2molal = .02 mol acetic acid / .1 kg vinegar.  .02 x 60.052 = 1.20104g = 1.20104 mL.

Is this correct (or close)?

The only other way I can think of doing it is to disregard molality too and just assume that 5% acetic acid = 100gx.05 = 5mL.

#### opti384

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #7 on: August 28, 2010, 10:54:50 PM »
I think you are getting on the right track. First of all, I don't think plugging in the mass to molality is correct. Rather you should consider the percent concentration (as the problem gives 5% acetic acid).

#### WhoCares357

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #8 on: August 28, 2010, 11:08:27 PM »
I think you are getting on the right track. First of all, I don't think plugging in the mass to molality is correct. Rather you should consider the percent concentration (as the problem gives 5% acetic acid).

So there are 5g of acetic acid. I'm tempted to just convert it into mL and accept that as answer. However, I feel this is wrong somehow. What was the point of telling me the molality and molarity?

I think my problem is that I don't know why I can't use molarity/molality in the beginning of the problem.

Isn't it true that 1M = 1mol/1L = .1mol/.1L = .5mol/.5L? Why couldn't I use it to find the moles in the first place? Does it only apply specifically when there is 1 L of solution or 1 mol of solute?

If I use the 5g answer to find the number of moles per .1L I get about .08 moles. If I plug .08 moles into the molarity equation I get a number other than .1L for the volume of the solution.

That's a bit of a long rant, but I think my question is how flexible is the molarity/molality equation? What does it even mean in this context?

#### opti384

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #9 on: August 28, 2010, 11:16:29 PM »
First of all by saying 5%, it does not mean that there are 5g of acetic acid. The percentage concentration is the "mass of solute/the mass of solution(solute + solvent)" in percentage.

You are right that 1M = 1mol/1L and so one. But here you should figure out what you are asked to find. The problem is that you take some from the 1M acetic acid and some from 0.2m acetic acid to make 100ml of 5% acetic acid. I think were confused because you were not familiar with the 5% concentration. So I recommend that you set an equation expressing for example how much mols of acetic acid are from 1M and how much mols of acetic acid are from 0.2 m acetic acid equals to mols of acetic acid in 100ml vinegar.

#### WhoCares357

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #10 on: August 28, 2010, 11:30:18 PM »
I don't understand what you mean by the example. Can you show me how you would set it up?

#### opti384

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #11 on: August 28, 2010, 11:36:18 PM »
Let's say we take x ml from 1M of acetic acid and y ml from 0.2m acetic acid. Then what should be the sum of the mols of acetic acid and sum of x and y?

#### WhoCares357

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #12 on: August 29, 2010, 12:01:20 AM »
I'm going to go get some sleep. I'll *Ignore me, I am impatient* this tomorrow if I still don't get it.

Thanks a lot for your help.

#### WhoCares357

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #13 on: August 29, 2010, 09:33:28 AM »
Let's say we take x ml from 1M of acetic acid and y ml from 0.2m acetic acid. Then what should be the sum of the mols of acetic acid and sum of x and y?

I still don't really understand how you're taking the mL. Are you just assuming the x is the mL of solute derived by using the molarity equation and y is the mL of solute derived by using the molality equation?

If I do it this way I get the equation: .1mol(60.052g/mol)/(1g/mL) + .02mol(60.052g/mol)/(1g/mL) = x + y = 7.20624.

Am I on the right track?

#### opti384

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##### Re: Conversion (molarity, molality, %, etc)
« Reply #14 on: August 29, 2010, 09:46:42 AM »
Quote
I still don't really understand how you're taking the mL. Are you just assuming the x is the mL of solute derived by using the molarity equation and y is the mL of solute derived by using the molality equation?

Yes. But you will not be able to directly find the mols of acetic acid in 0.2 molal solution. I recommend that you label the mass of solvent and solute for each 1M acetic acid, 0.2 molal, and 100ml vinegar. And also I made a mistake: there are 5g of acetic acid in 100ml vinegar, my bad.