Final answer:
The problem is a mathematical exercise in mixing solutions of different acid concentrations to achieve a desired acid percentage in the final mixture. We set up a system of equations to represent the volume constraints and the percentages of acid in each solution. By solving these equations simultaneously, we determine the volumes of Solution A, B, and C used in the mixture.
Step-by-step explanation:
The problem is a mathematical exercise in algebra, specifically dealing with the method of mixing solutions of different concentrations to achieve a mixture with a desired concentration of solute. The problem involves creating a mixture from three solutions with varying percentages of acid. In this case, the solutions contain 15%, 5%, and 40% acid respectively. When combined, the chemist aims to have a 60-liter mixture with a 29% acid concentration. Additionally, the problem states that Solution C is used twice as much as Solution A in the mixture.
To solve this, we use a system of equations:
- The total volume of the mixture is 60 liters: A + B + C = 60.
- Twice as much Solution C as Solution A is used: C = 2A.
- The total amount of acid in the final solution is 29% of 60 liters: 0.15A + 0.05B + 0.40C = 0.29 × 60.
Substituting C = 2A into the first two equations, we can solve for A and B. Substituting A and B back into the first equation, we can then solve for C. The quantities of A, B, and C that satisfy all three equations give us the volume of each Solution used in the mixture.
Example Calculation:
Let's assume A is the volume of Solution A used in liters. Since C = 2A, the third equation becomes 0.15A + 0.05B + 0.40(2A) = 17.4 (since 0.29 × 60 = 17.4). This simplifies to 0.95A + 0.05B = 17.4. From the first equation A + B + 2A = 60, we get 3A + B = 60. Now, we have two equations and two unknowns which we can solve simultaneously to get the volumes of A and B, and subsequently, C.