Question

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

Answer 1

This means that you need to add 1 gallon of water to 1 gallon of pure antifreeze to obtain a solution that is 75% antifreeze. To obtain a solution that is 75% antifreeze, you need to add 1 gallon of water to 1 gallon of pure antifreeze.

Step-by-step explanation:

To obtain a solution that is 75% antifreeze, you need to determine how much water should be added to 1 gallon of pure antifreeze.

Let's assume that the final volume of the solution, after adding water, remains at 1 gallon.

The amount of antifreeze in the solution can be calculated using the equation:

Amount of antifreeze = Volume of antifreeze / Total volume of solution

Since the volume of the antifreeze is 1 gallon (given) and the total volume of the solution is also 1 gallon (assuming), we can calculate:

Amount of antifreeze = 1 gallon / 1 gallon = 1

This means that you need to add 1 gallon of water to 1 gallon of pure antifreeze to obtain a solution that is 75% antifreeze.

Answer 2

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