How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 75 %75% antifreeze?

For the purpose we will use solution dilution equation:

c1V1=c2V2

Where, c1 - concentration of stock solution; V1 - volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.

c1 = 100%

c2 = 75%

V1 = 1 gal

V2 = ?

When we plug values into the equation, we get following:

100 x 1 = 75 x V2

V2= 100/75 = 1.33 gal

Now we can determine the necessary volume of water:

V(water)=  V2 - V1 = 1.33 - 1 = 0.33 gal water should be added