7% glacial acetic acid in 40% (v/v) methanol

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To prepare a 7% glacial acetic acid solution in 40% (v/v) methanol, clarity on the definitions of w/w%, v/v%, and w/v% is essential. The confusion arises from whether to mix 7 mL of acetic acid with 40 mL of methanol and then dilute to 100 mL, or to create a 7% acetic acid solution first and then combine it with methanol. The first approach is considered more reasonable, with a suggestion to start with 40 mL of methanol, add 60 mL of water for the 40% v/v methanol, and then add approximately 7.3 mL of glacial acetic acid to account for volume contraction. The ambiguity in the original document adds to the confusion, but the discussion emphasizes the importance of understanding solution preparation. Further questions are anticipated as users seek clarity on similar topics.
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How do I make 7% glacial acetic acid in 40% (v/v) methanol?
 
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7% w/w, 7% w/v or 7% v/v?

Please note you should show your attempts at solving the question to receive help. This is a forum policy.
 
I am totally confused, I don't understand how to solve it. :(
Please see also: http://www.sigmaaldrich.com/etc/medialib/docs/Sigma/Datasheet/b2025dat.Par.0001.File.tmp/b2025dat.pdf

I need to make this solution. Also I need to make 10% acetic acid in 25% (v/v) methanol.
 
7% in 40%v/v is ambiguous, no wonder you are confused.

Do you at least know definitions of w/w%, v/v% and w/v%?
 
I think I know the definition for those.
Do you think I have to make 7 ml(acetic acid) + 40 ml(methanol) and add water up to 100 ml?

Or Do I have to make 7% acetic acid (7ml in 100ml of water) and then add it to 40% methanol? If I do the 2nd way, how much of acetic acid I need to add to Methanol?

Or Is these are not right & I have to prepare it in a different way?
 
First approach sounds more reasonable, although I would go slightly different way - I would start with 40 mL of methanol, add 60 mL of water (that would give 40%v/v methanol), then add 7.3 mL of glacial acetic acid (0.3 mL more to account for the fact final volume would be around 104 mL, as volume of the methanol solution would be around 97 mL due to the volume contraction). At least that's how I understand the composition of the solution. But as I wrote earlier, document you have linked to is ambiguous, so could be I am wrong.
 
I believe what you suggested is more appropriate for this context.

Thanks a lot.

At first, I thought I would just write the question and get the answer. But you made me think. I appreciate your policy. Thanks again.
I am sure I will have a lot more questions to ask. :)
 
dr.vj said:
I am sure I will have a lot more questions to ask.

Be prepared for more thinking then.
 
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