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Rubbing alcohol contains 70% isopropyl alcohol. Suppose you are adding 180ml of liquid that contains 45% isopropyl alcohol. How much liquid should be added of the higher concentration to obtain rubbing alcohol that is 70% isopropyl alcohol?

User Alex Riina
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Starting with some solution of volume v and an alcohol concentration of 100c % , if you add 180 mL of 45% isopropyl to it, you get a mixture with a total volume of 180 mL + v.

Each mL of the starting solution contains c mL of alcohol. For example, if the concentration of the starting solution is 80% (so c = 0.8), then each mL contains 80% = 0.8 of one mL of alcohol.

Similarly, each mL of the added solution with 45% concentration contains 0.45 mL of alcohol.

You want the new solution to have a concentration of 70%, so that the ratio of the amount of alcohol in it to the total volume is 70%, meaning

cv + 0.45 (180 mL) = 0.7 (180 mL + v)

and you want to solve for v :

cv + 81 mL = 126 mL + 0.7v

cv - 0.7v = 126 mL - 81 mL

(c - 0.7) v = 45 mL

v = (45 mL) / (c - 0.7)

Judging by context clues, c lies somewhere between 0.7 and 1. (It can't be less than 0.7 because mixing this solution with any other solution of smaller concentration will never yield a solution of higher concentration.)

If, for example, c = 0.8, so that your starting solution is at 80% concentration, then you would need

v = (45 mL) / (0.8 - 0.7) = 450 mL

to be mixed with 180 mL of 45% solution to end up with 180 mL + 450 mL = 630 mL of 70% solution.

As another example, if you're mixing pure alcohol, so that c = 1, you would need

v = (45 mL) / (1 - 0.7) = 150 mL

to make a 180 mL + 150 mL = 330 mL batch of 70% solution. The fact that you would need less of a higher concentration solution is not surprising.

User Kagundajm
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