How many liters of a 10 percent alcohol solution must be mixed with 80 liters of a 80 percent solution to get a 50 percent solution?

Answer 1

The following solution is just an approximation, in practical circumstances sufficient to use. The addendum explains why.

• x = number of litres (10%) to be added. You have 64 litres of alcohol in your original 80 litres.
• 64 + x/10 = (80 + x)/2
• 640 + x = 400 + 5x
• 4x = 240
• x = 60 and alcohol content (10%) will be 6.
• so you will have a total of 140 litres of which 64 + 6 will be alcohol.

Added note

to deal with the so called 'dilution contraction' of total volume

If it were % by MASS ( %m/m), it's quite easy to do (based on the 'Mass Conservation Law). You calculate with mass (kg) and mass-% (%m/m) i.s.o. volume (L) and vol% (%v/v).

However if the meaning was: % by Volume ( %v/v) then calculation appears to become quite complicated, but not impossible if you know at least the density values of all solutions (original 80%v/v or 10%v/v and final 50%v/v).

DO NOT use: (orig. volume) + (added volume) = final volume, as done above, if exact figures are necessary.

It's only a rule of thump, an approximation. This is because fluids can contract on mixing at dilution. There is no rule such as: conservation of volume.

Your case: 60 L + 80 L (is not equal but) < 140 L final solution.