Professor Beaker has a 50-milliliter solution containing 60% acid. How many milliliters of pure water should he add to this solution to produce a solution containing 40% acid?
see related questions below.
I've tried looking at similar problems but I'm still confused.
the amount of acid in the parts must add up to the amount in the final mix. If x ml of water (0% acid) are added, then you need
.60*50 + 0x = .40(50+x)
Now just solve for x. This assume you know what "%" means ...
To solve this problem, we need to find out how many milliliters of pure water Professor Beaker should add to the 50-milliliter solution containing 60% acid in order to obtain a solution with 40% acid.
First, let's calculate the amount of acid in the original solution. The solution contains 60% acid, which means that 60% of the 50 milliliters is acid. We can find this value by multiplying 50 by 0.6:
Amount of acid in the original solution = 50 mL × 0.6 = 30 mL
Now, we need to find out how much total volume the final solution will have. Since we are adding pure water to the original solution, the total volume will be the sum of the original solution volume and the volume of water added. Let's call the volume of water added "x" milliliters.
Total volume of the final solution = 50 mL + x mL
The final solution is required to have 40% acid. To calculate the amount of acid in the final solution, we multiply the total volume of the final solution by 0.4:
Amount of acid in the final solution = (50 mL + x mL) × 0.4
Since the acid is not added or removed when mixing solutions, the amount of acid in the original solution and the final solution should be the same. Therefore, we can equate the amount of acid in the original solution to the amount of acid in the final solution and solve for x:
30 mL = (50 mL + x mL) × 0.4
Now, we can solve this equation for x:
30 mL = (0.4) × (50 mL + x mL)
30 mL = 20 mL + 0.4x mL
30 mL - 20 mL = 0.4x mL
10 mL = 0.4x mL
10 mL / 0.4 = x mL
25 mL = x mL
Therefore, Professor Beaker should add 25 milliliters of pure water to the 50-milliliter solution to produce a solution containing 40% acid.