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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.
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
