I have 945ml of 75% ethanol solution?
Answer:
in most biomed labs, 75% ethol means by volume, for convenience. its always a rough guide anyway, you don't need too much precision because the longer you take the more ethol evaporates.
are you trying to bulid a meth lab..good luck not blowing yourself up
You need to do this in mass or mole units, trust me. When you add volumes of liquids, the total volume has a habit of NOT adding to the sums of the components (partial molar volumes). If I were you, I would refuse to work with any numbers that weren't in mass.
What I'm saying is that 50% is not descriptive enough for me to answer your question. If that is a volume percent, you're on your own...
You haven't mentioned if you wanted percentages by volume or by weight. The procedures are slightly different for each, but it is possible to do.
For the "by volume" approach, you need 500 L of ethanol. With your on-hand solution, you would need 667 ml [(4/3)*500] , and dilute with 333 ml of water. For the "by weight" approach, you need to know the density of the 75% solution and a 50% solution; such info is available. Then you can figure the amount of solution that provides the ethanol that would occur in one 1 liter of a 50% solution.
Easy parcheesy... This is a simple C1V1 = C2V2 problem.
you have C1 = 75% solution
C2 = 50% solution (Concentration you want)
V2 = 1000ml (The volume you want)
Simply plug and chug.
(75)(V1) = (50)(1000)
V1 = 666.66 ml of your 75% ethanol solution
add this volume to 333.33 ml H20 and you have a 50% ethanol solution.
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