chemist has 6liter of 50% acid solution .how much water added to obtain 20% acid solution?
You want to dilute it from .5 to .2 or 5/2=2.5 times. So you add one part original stuff, then 1.5 parts water.
So add 9 liters water to get .2 percent acid.
To find out how much water needs to be added to obtain a 20% acid solution, we can use a basic algebraic approach. Here's how you can solve it step by step:
Step 1: Determine the amount of acid in the original 50% acid solution.
Since the chemist has a 6-liter solution that is 50% acid, we can calculate the amount of acid in the solution using the formula:
Acid = 6 liters × (50/100) = 3 liters.
Step 2: Set up the equation for the final 20% acid solution.
Let's assume we need to add "x" liters of water to the original solution. In the final solution, the amount of acid would still be 3 liters (as calculated in step 1), and the total volume would be 6 + x liters. Therefore, we can set up the equation:
3 liters = (20/100) × (6 + x) liters.
Step 3: Solve the equation to find the value of "x."
To solve the equation, we'll start by simplifying the right-hand side:
3 liters = (1/5) × (6 + x) liters.
Next, cross-multiply to eliminate the fractions:
3 × 5 × liters = 1 × (6 + x) liters.
Simplify further:
15 liters = 6 + x liters.
Now, isolate the value of "x" by subtracting 6 from both sides:
15 liters - 6 liters = x liters.
Thus, x = 9 liters.
So, the chemist needs to add 9 liters of water to the original 6-liter 50% acid solution to obtain a 20% acid solution.