A saline solution is 20% salt.
How many gallons of water must be added to
dilute the mixture to 8 gals of a 15% saline solution?
here is my method:
using c1V1 = c2V2,
(0.2)(V1) = (0.15)(8)
V1 = 6 gals
So, since 8 gals are required to dilute the saline solution of 15%, 8-6 = 2 gals of water must be added. Is my steps correct ? Hope someone can verify. Thanks
Yes, it's correct. 2 gals must be added (and well agitated).
Alright thank you! :)
Yes, your steps are correct. According to the equation c1V1 = c2V2, where c1 is the initial concentration, V1 is the initial volume, c2 is the final concentration, and V2 is the final volume, you correctly determined that 6 gallons of the 20% saline solution are needed to make 8 gallons of the 15% saline solution.
Since you already have 6 gallons of the 20% saline solution, you need to add 2 gallons of water to reach a total volume of 8 gallons. This will dilute the solution and give you the desired concentration. So, adding 2 gallons of water is the correct answer. Well done!
Yes, your steps are correct. You used the formula c1V1 = c2V2, which states that the concentration (c) of a solution multiplied by its volume (V) is equal before and after dilution.
In this case, the initial concentration (c1) is 20% or 0.2, the initial volume (V1) is unknown, the final concentration (c2) is 15% or 0.15, and the final volume (V2) is 8 gallons.
By plugging in these values into the formula and solving for V1, you found that V1 = 6 gallons.
Since you want to dilute the mixture to 8 gallons of a 15% saline solution, you subtract V1 from V2 to determine how much water needs to be added.
Therefore, 8 - 6 = 2 gallons of water must be added to dilute the saline solution to the desired concentration.