how many liters of distilled water must be added to 80 liters of 60% acid solution to obtain a 50 % acid solution?
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To solve this problem, we need to determine the amount of distilled water that should be added to the 80 liters of 60% acid solution.
To start, let's define the variables:
Let "V" represent the volume of distilled water we need to add (in liters).
Now, let's set up the equation based on the acid content in the initial and final solutions:
The acid content in the initial solution can be calculated as follows:
Acid content in the initial solution = 80 liters × 60% = 48 liters
The acid content in the final solution can be calculated as follows:
Acid content in the final solution = (80 liters + V) × 50%
According to the conservation of acid content, the acid content in the initial and final solutions must be equal. Therefore, we can set up the equation:
48 liters = (80 liters + V) × 50%
To solve this equation, we need to isolate the variable "V".
First, let's divide both sides of the equation by 50% (or 0.50) to get rid of the percentage:
48 liters ÷ 0.50 = 80 liters + V
Simplifying,
96 liters = 80 liters + V
Now, let's isolate V by subtracting 80 liters from both sides of the equation:
96 liters − 80 liters = V
V = 16 liters
Therefore, you need to add 16 liters of distilled water to the 80 liters of 60% acid solution to obtain a 50% acid solution.
amount added ..... x L
solve for x,
.6(80) + 0.0x = .5(80+x)
48 = 40 + .5x
x = 16