You have 200 proof (100%) ethanol, but you need 50 mL of 70% ethanol. What would

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Use this formula M1*V1=M2*V2, where M1 = Your original concentration --> 100% M2 = Your desired concentration --> 70% V1 = Your original volume of solution (You'll need to calculate this) V2- Your desired concentration --> 50mL Okay lets pop in the numbers into the formula: M1V1:V2M2 and solve for V1. (100%)(V1)=(50mL)(70%) V1 = (50mL)(70%) / (100%) = 35mL V1 = 35mL The reason you're trying to solve for V1 (35mL) is so you'll know how much to go get from the stock solution. You can imagine the stock solution is infinitally plentiful, so you'll need to know the minimum amount you'll need. Imagine diluting the entire stock solution of 100% ethanol down to 70% ethanol. We'll be dumping buckets of DI water all day just to get an olympic sized pool of 70% ethanol! And we only needed 50mL!! hahah I'm only exagreating of course. :) Okay so heres what you do: You'll take a 1L graduated cylinder, pour 35mL of 100% ethanol into the graduated cylinder. Then you'll get some DI (deionized) water and pour into the graduated cylinder with the 100% ethanol until the meniscus touches the 50mL mark on the graduated cylinder. [So basically if you do the math, you add 15 mL of DI water in 35mL of 100% ethanol.] Adding 15 mL of DI water over 35mL of 100% ethanol, dilutes the 100% ethanol down to 70% ethanol. You ever heard of a watered-down liquor, its the same idea! Not only that, but the 70% concentration just so happens to be the desired concentration of 50mL you were asked for. Notes: If you want to be more precise with your measurement, use a Grade A volumetric 50mL cylinder. Okay.. I hope I earned a lifesaver!! Hope this helps everybody

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