June 04, 2020, 02:13:04 AM
Forum Rules: Read This Before Posting


Topic: Conversion (molarity, molality, %, etc)  (Read 12444 times)

0 Members and 1 Guest are viewing this topic.

Offline WhoCares357

  • Regular Member
  • ***
  • Posts: 21
  • Mole Snacks: +0/-0
Re: Conversion (molarity, molality, %, etc)
« Reply #15 on: August 29, 2010, 10:08:55 AM »
Okay so the solution is .1L or .1kg. The solute is 5g. The solvent is 95g.

Molarity = moles of solute / liters of solution, molality = moles of solute / liters of solvent

I find the moles from the molarity (1M = .1mol/.1L solution) as .1 mol.
I find the moles from the molality (.2molal = .019 mol / 95g solvent).

Then I find the concentration based on the two moles that I got.
x = .1mol(60.05g/mol) = 6.005g
y = .019mol(60.05g/mol) = 1.141g

Now that I have the concentration what do I do with it?

Offline opti384

  • Full Member
  • ****
  • Posts: 434
  • Mole Snacks: +33/-25
  • Gender: Male
    • In the Search for the Laws of Nature
Re: Conversion (molarity, molality, %, etc)
« Reply #16 on: August 29, 2010, 10:12:23 AM »
I find the moles from the molality (.2molal = .019 mol / 95g solvent).

Be careful. Is the solvent of 95g of vinegar all from 0.2m of acetic acid?

Offline WhoCares357

  • Regular Member
  • ***
  • Posts: 21
  • Mole Snacks: +0/-0
Re: Conversion (molarity, molality, %, etc)
« Reply #17 on: August 29, 2010, 10:27:18 AM »
Why wouldn't it be? I thought you could scale the mole/mass in the equation.

Offline opti384

  • Full Member
  • ****
  • Posts: 434
  • Mole Snacks: +33/-25
  • Gender: Male
    • In the Search for the Laws of Nature
Re: Conversion (molarity, molality, %, etc)
« Reply #18 on: August 29, 2010, 10:29:41 AM »
But the problem states that you take some ml from 1M acetic acid and some from 0.2m acetic acid. Therefore the solvent of 95g will be sum of the mass of the solvent take from 1M and 0.2m.

Offline WhoCares357

  • Regular Member
  • ***
  • Posts: 21
  • Mole Snacks: +0/-0
Re: Conversion (molarity, molality, %, etc)
« Reply #19 on: August 29, 2010, 10:38:42 AM »
So 95 = xg from molarity + .2xg from molality?

If I do that then I get .084 L of solution (.079 solvent + .005 solute) which is .084 mol from Molarity and .016 L which is .0032 mol from molality. Converting the moles to grams I get 5.04 and .192

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +467/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: Conversion (molarity, molality, %, etc)
« Reply #20 on: August 29, 2010, 12:27:24 PM »
I'd start by converting all concentrations to M (=mol/L).

You can do this easily as the question states that you may assume all solutions have density 1 g/mL.

You can express the question in terms of two simultaneous equations:

V1M1 + V2M2 = V3M3 (1)
V1 + V2 = V3 (2)

Where:

V# = Volume of solution #
M# = Molarity of solution #

We will call the 1 M solution "solution 1", the 0.2 m solution "solution 2" and the 5% solution "solution 3".

We already know V3, that's 100 mL (= 0.1 L), and M1, that's 1 M.

So to solve the equations we need to:

1. Calculate the molarity M2 - ie. convert 0.2 mol/Kg into mol/L [hint: assuming density of all solutions is 1 g/mL]

2. Calculate M3. [hint: you have already worked out that there is 5 g acetic acid in 100 mL of solution 3 - convert this information into M (=mol/L)]

3. Now the only unknowns are V1 and V2. Two unknowns in two simultaneous equations (1) and (2), solve these equations for V1 and V2 and your homework is done.
My research: Google Scholar and Researchgate

Offline WhoCares357

  • Regular Member
  • ***
  • Posts: 21
  • Mole Snacks: +0/-0
Re: Conversion (molarity, molality, %, etc)
« Reply #21 on: August 29, 2010, 12:51:44 PM »
So my answer is V1 + V2?

Now that I understand the process clearly I have a question about it. Why isn't the answer just 5mL (from the V3M3)? If I had to guess I would say that the number of moles is different between doing that and using the molarity equation. But if that is true, why is it true? Does molarity/molality always apply to a specific volume 0f a solution only?

Thank you very much for the explanation. I'm just trying to understand why you did what you did. I probably don't know enough about the different concentrations to figure it out.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 25779
  • Mole Snacks: +1686/-400
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Conversion (molarity, molality, %, etc)
« Reply #22 on: August 29, 2010, 03:23:58 PM »
Somehow I doubt you understand what it is all about.

For now let's forget about different ways of expressing concentrations. They can be always converted one to another, so to some extent it doesn't matter what units the concentration is in.

Question is about mixing two solutions. When you mix them, amount of solute in the final solution will be sums of amounts of solute in both solutions, and amount of solvent will be sum of amount of solvents in both solutions. This is based on one of the most fundamental laws - law of mass conservation.

Is that clear?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline WhoCares357

  • Regular Member
  • ***
  • Posts: 21
  • Mole Snacks: +0/-0
Re: Conversion (molarity, molality, %, etc)
« Reply #23 on: August 29, 2010, 04:16:37 PM »
I think I understand. I just assumed that the 1M and the .2molal were interchangeable. I didn't realize it was describing two different parts of the solute.

Thank you for the explanation.

Sponsored Links