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Imagine you have two ingredients - 45 grams of 60% ethanol, and as much water as you wish. The only accurate tool you have is a scale, which measuring error is up to $\pm \pu{1gram}$. How would you create five different solutions of 1, 2, 3, 4, and 5% ethanol?

It can be a solution of % ethanol by volume or by weight, depending on which can be created more accurately with the tools at hand. It might be super easy to you, but as an electrical engineer the logic and math eludes me.

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  • $\begingroup$ Have you seen the equation C1V1=C2V2 before? $\endgroup$ – David Wyn Williams May 16 '18 at 10:02
  • $\begingroup$ Nope - What is it called and what does it refer to? $\endgroup$ – Mike Meyers May 16 '18 at 10:03
  • $\begingroup$ It's used to calculate the concentration of your solution after dilution. C1 is your initial concentration, C2 is your final concentration (e.g 5%) and V2 is your final volume (E.g 100 ml). V1 is what you calculate. mathcentre.ac.uk/resources/uploaded/c1v1c2v2.pdf $\endgroup$ – David Wyn Williams May 16 '18 at 10:22
  • $\begingroup$ Would it work even though the 60% ethanol water mixture's density is about 0.9? wissen.science-and-fun.de/chemistry/chemistry/density-tables/… $\endgroup$ – Mike Meyers May 16 '18 at 10:48
  • $\begingroup$ Yep. So if you have 60% ethanol and need 5% solution, v1=(5% x 100 m l) / 60% Which means you need 8.33 ml of 60% ethanol in 100 ml solution. (8.33 ml ethanol, 91.77ml of water) $\endgroup$ – David Wyn Williams May 16 '18 at 11:39
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You should use the concentration dilution equation

$$C_1V_1 = C_2\cdot V_2$$ If you re-arrange it so the volume of 60% ethanol needed is:

$$V_1 = \frac{C_2\cdot V_2}{C_1}$$

You can then insert all the values you do know and there you go.
For example, if you need a $\pu{100 m}$ of 5% solution: $$V_1=\frac{5\% × \pu{ 100 ml}}{60\%}$$

Which means you need $\pu{8.33 ml}$ of 60% ethanol in $\pu{100 ml}$ solution.

As max pointed out, since you only have a scale, use d=m/v. D = density, m = mass and v= volume You re-arrange so mass = d.v

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  • $\begingroup$ That doesn't work if all you have is a scale that weighs +/- 1 gram as the OP posed. $\endgroup$ – MaxW May 16 '18 at 14:48

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