I need to make a solution that contains PAH (paraaminohippuric acid) at 25mmol/liter.
is this procedure right?
1. 25mmol/l divide by 1000=0.025mol/l
2. 0.025 mol/l x the MW of PAH which is 194.19g/mol =4.85475
does this mean i need to dissolve 4.85475 g of PAH in one liter to get the 25mmol/liter? What if the source of PAH is 99% or lower?
thanks in advance
nor
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Our forum has been available for many years. In September 2014 we decided to switch it to read-only mode. Month after month we saw less posts with questions and answers from real people and more spam posts. We were spending more and more resources cleaning the spam until there were less them 1 legitimate message per 100 spam posts. Then we decided it's time to stop.
All the posts in the forum will be available and searchable. We understand there are a lot of useful information and we aren't going to remove anything. As for the new questions, you can always ask them on convert-me.com FaceBook page
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get final concentration
by NOR » Tue Jul 18, 2006 3:03 am
- NOR
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Re: get final concentration
by Guest » Tue Jul 18, 2006 6:57 pm
NOR wrote:I need to make a solution that contains PAH (paraaminohippuric acid) at 25mmol/liter.
is this procedure right?
1. 25mmol/l divide by 1000=0.025mol/l
2. 0.025 mol/l x the MW of PAH which is 194.19g/mol =4.85475
does this mean i need to dissolve 4.85475 g of PAH in one liter to get the 25mmol/liter?
Yes. You will need about one liter of water, but volume may not be perfectly additive. Add water until you have 1 L of solution.
What if the source of PAH is 99% or lower?
thanks in advance
nor
Express the percentage as a decimal fraction (0.99). Divide the grams above by the purity fraction.
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- NOR
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