Question

8. Three solutions contain a certain acid. Solution A contains 15% acid.

Solution B contains 5% acid, and Solution C contains 40% acid. A chemist

combined all three solutions to create a 60-liter mixture containing 29%

acid. If twice as much of Solution C as Solution A was used, how many

liters of each solution was used for the mixture?

Answer 1

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:

1. The total volume of the mixture is 60 liters: A + B + C = 60.
2. Twice as much Solution C as Solution A is used: C = 2A.
3. 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.