how many liters of water must be added to 30 liters of a 20% alcohol solution to dilute it to a 15% solution?
you are diluteing it 20/15=1.333 times
Add one part solution, .333333 water.
This means if one part is 30 liters, add 10 liters water.
Check: .2*30/(40+10)=6/40=.15 or 15 percent.
To solve this problem, we need to use the concept of mixtures and the principles of concentration.
Let's break down the problem into steps:
Step 1: Determine the initial amount of alcohol in the solution.
In the 30 liters of 20% alcohol solution, we have 20% of 30 liters as alcohol.
20% of 30 liters = (20/100) * 30 = 6 liters of alcohol.
Step 2: Determine the final concentration.
We want to dilute the solution to a 15% concentration. This means that the final solution should contain 15% alcohol.
Step 3: Set up the equation.
Let's assume we add 'x' liters of water to dilute the 30 liters of 20% alcohol solution.
The amount of alcohol in the final solution would be the sum of the amount of alcohol in the original solution and the added water. Since water does not contain alcohol, the amount of alcohol remains the same.
So, the equation is: 6 liters (initial alcohol) = 15% (final concentration) * (30 liters + x liters).
Step 4: Solve the equation.
6 = 0.15 * (30 + x).
Step 5: Solve for 'x'.
Divide both sides of the equation by 0.15:
6 / 0.15 = 30 + x.
40 = 30 + x.
Subtract 30 from both sides of the equation:
40 - 30 = x.
x = 10 liters.
Therefore, to dilute the 30 liters of 20% alcohol solution to a 15% solution, you need to add 10 liters of water.