A teaching assistant at a university needs 28% acid solution for her class's lab experiment. There isn't any of this concentration in stock, but the lab has 50 liters of 22% acid solution, as well as a lot of 32% acid solution. How much of the 32% acid solution should the teaching assistant add to the 22% acid solution to obtain a solution with the desired concentration? Write your answer as a whole number or as a decimal rounded to the nearest tenth.____ liters
Q. A teaching assistant at a university needs 28% acid solution for her class's lab experiment. There isn't any of this concentration in stock, but the lab has 50 liters of 22% acid solution, as well as a lot of 32% acid solution. How much of the 32% acid solution should the teaching assistant add to the 22% acid solution to obtain a solution with the desired concentration? Write your answer as a whole number or as a decimal rounded to the nearest tenth.____ liters
Denote Acid Solution Volume: Let's denote the amount of 32% acid solution that needs to be added as x liters. The total volume of the new solution will be 50 liters + x liters.
Calculate Pure Acid in 22% Solution: The amount of pure acid in the 50 liters of 22% solution is 0.22×50 liters.Calculation: 0.22×50=11 liters of pure acid.
Calculate Pure Acid in 32% Solution: The amount of pure acid in the x liters of 32% solution is 0.32×x liters.Calculation: 0.32×x=0.32x liters of pure acid.
Calculate Total Pure Acid: The total amount of pure acid in the final solution should be 28% of the total volume, which is 28% of (50+x) liters.Calculation: 0.28×(50+x)=14+0.28x liters of pure acid.
Set Up Equation: To find the value of x, we set up the equation that the sum of the pure acid from both solutions equals the pure acid in the final solution: 11+0.32x=14+0.28x.
Solve for x: Now we solve for x:0.32x−0.28x=14−11,0.04x=3,x=0.043,x=75 liters.
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