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Topic: finding molaritay and other concentrations  (Read 9822 times)

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Offline tinoco19

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finding molaritay and other concentrations
« on: September 01, 2013, 02:01:20 AM »
What percent solution is a 0.69M solution of Acetic acid? If 1M of Acetic Acid = 60 grams, I divided .69 from 60 and found that the solution would be 86.96%.  Am I on the right track?
« Last Edit: September 01, 2013, 04:18:57 AM by tinoco19 »
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Offline tinoco19

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Re: finding molaritay and other concentrations
« Reply #1 on: September 01, 2013, 02:04:10 AM »
What is the weight/volume of a 32% (mass to volume) solution of Acetic Acid?
edited-sorry didn't realize the problem had to be attempted...here's what i got: Acetic Acid is 60 grams. 32% of the solution means there's 32 g/100ml.  Would 32g/100ml be the final answer
« Last Edit: September 01, 2013, 04:38:10 AM by tinoco19 »
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Offline tinoco19

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Re: finding molaritay and other concentrations
« Reply #2 on: September 01, 2013, 02:07:36 AM »
What is the molarity of a 32% solution of acetic acid?
modified-again sorry I didn't realize I had to post my attempts...
I concluded that acetic acid is 60 g/L = 1M.  A 32% solution of Acetic Acid would mean there's 32g/100ml, so to find the answer...32g/60g =.53  Is that correct?
« Last Edit: September 01, 2013, 04:22:09 AM by tinoco19 »
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Offline Archer

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Re: finding molaritay and other concentrations
« Reply #3 on: September 01, 2013, 02:14:11 AM »
You must attempt the problem, this is a forum rule.

 http://www.chemicalforums.com/index.php?topic=65859.0
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Offline Archer

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Re: finding molaritay and other concentrations
« Reply #4 on: September 01, 2013, 02:40:44 AM »
Quote from: tinoco19 on September 01, 2013, 02:01:20 AM
What percent solution is a 0.69 solution of Acetic acid?

0.69 molar?
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Offline tinoco19

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Re: finding molaritay and other concentrations
« Reply #5 on: September 01, 2013, 03:22:58 AM »
Quote from: Archer on September 01, 2013, 02:14:11 AM
You must attempt the problem, this is a forum rule.

 http://www.chemicalforums.com/index.php?topic=65859.0

thank you! sorry about that.  I edited to show in my attempted tries.
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Offline Archer

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Re: finding molaritay and other concentrations
« Reply #6 on: September 01, 2013, 03:31:25 AM »
Quote from: tinoco19 on September 01, 2013, 02:01:20 AM
What percent solution is a 0.69M solution of Acetic acid? If 1M of Acetic Acid = 48 grams, I divded .69 from 48 and found that the solution would be 33.12%.  Am I on the right track?

What is the molecular weight of acetic acid? Please include units in your answer.
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Offline tinoco19

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Re: finding molaritay and other concentrations
« Reply #7 on: September 01, 2013, 04:17:42 AM »
Molecular weight of Acetic Acid (C2H4O2) would be 60 grams....not 48 as I previously thought. I must've added wrong.
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Offline Archer

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Re: finding molaritay and other concentrations
« Reply #8 on: September 01, 2013, 06:44:45 AM »
Quote from: tinoco19 on September 01, 2013, 04:17:42 AM
Molecular weight of Acetic Acid (C2H4O2) would be 60 grams....not 48 as I previously thought. I must've added wrong.

Molecular weight is not measured in grams.

What are the units of molecular weight?
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Offline magician4

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Re: finding molaritay and other concentrations
« Reply #9 on: September 01, 2013, 11:57:49 AM »
if I don't miss something here, "% solution" would mean something like "grams of acetic acid per gram of solution thereof"

starting from c = 0.69 mol/L (which is not "next to nothing"), for precise results we'd also have to consider the density of the system at a given temperature - or else have to explicitly state that as an approximation we're calculating with 1 g/cm³


regards

Ingo
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Offline Borek

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Re: finding molaritay and other concentrations
« Reply #10 on: September 01, 2013, 02:31:56 PM »
1.0043 g/mL, safe to ignore.
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