I have to prepare a 0.2% water solution of n-propanol and a water solution of methanol and isopropanol at 0.2% that can be in the same container. It didn't specify but I'm pretty sure the solutions are v/v%.
For the n-propanol a 0.2% solution should just be 2mg/mL of solution and the mg can be converted using the density to calculate the volume of n-propanol required, that I understand. What confuses me though is the second solution which is methanol and isopropanol in the same container, do I put 1mg worth of both methanol and ispropanol (after converting) in a mL of water OR do I make two separate 0.2% solutions of each and mix them together.
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If you interpret the question to be to prepare 0.2% solution methanol and 0.2% solution of isopropanol, and you want to put them in the same container, you add the volume of each required as if the other component were not there. Remember that if you prepare them separately and mix them that each will dilute the other when the final solution is made.
Oh that's true, so you're saying to measure out the amount of methanol needed for a 0.2% solution and the amount of isopropanol needed for a 0.2% solution, adding both to the same 1 mL of water, correct?
yes
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