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Author: Subject: Noob Solution Percentage Question
arkallic
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Noob Solution Percentage Question

Hey everyone.

So this is kind of a noob question.

I'm gonna be doing a project that calls for "10ml of a 10% solution of KOH in 95% ethanol"

So here's my question. What's the right way to figure out how much KOH to put in the 10ml of ethanol?

If the ethanol is 7.89g, do I just take 10% of that (.79g)?

Or do I go by moles?

I did the math in moles and got about .94g of KOH.

So what way is right? Or with such a small difference does it even matter?

[Edited on 31-1-2014 by arkallic]
DraconicAcid
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The 10% is almost certainly by mass (although some chemists would use mass/volume percent, which is easy, yet ridiculous), never moles. That would give you 0.8 g of KOH, or 1 g KOH if it's mass/vol%. In most applications, it probably won't matter, because your potassium hydroxide isn't pure anyway (commercial KOH is probably 96-98% KOH, with the rest being potassium carbonate and water)

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arkallic
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Thanks very much!
blogfast25
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 Quote: Originally posted by DraconicAcid , because your potassium hydroxide isn't pure anyway (commercial KOH is probably 96-98% KOH, with the rest being potassium carbonate and water)

Actually, commercial KOH almost always contains 10 w% water.

 Sciencemadness Discussion Board » Fundamentals » Beginnings » Noob Solution Percentage Question Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Beginnings   » Responsible Practices   » Miscellaneous Engineering and Equipment   » Reagents and Apparatus Acquisition   » Electronics   » Process Engineering   » Materials and Metallurgy Special Topics   » Analytical Chemistry   » Electrochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques Literature and Documentation   » Sciencemadness Wiki   » Prepublication Non-science   » Forum Matters   » Legal and Societal Issues