How much pure alcohol must be added to 7 liters of 10 percent alcohol solution to increase the concentration to 30 percent alcohol?

Answer 1

2 liters

Let X represent the amount of pure alcohol to be added to the 7 liters of 10% alcohol solution to get a 30% alcohol solution. We want .3(7+x)=.1*7+x [because we need the alcohol in the final solution to equal the amount of alcohol in the original solution plus the amount of alcohol we are adding). Solving for x, we get:

.3(7+x)=.1*7+x

2.1+.3x=.7+x

1.4=.7x

2=x

Thus, 2 liters of pure alcohol needs to be added to the solution.

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 100%v/v or 10%v/v and final 30%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: 7 L + 2 L (is not equal but) < 9 L final solution.