A chemist has a solution that is 25% acid. How much of this solution must be mixed with 10 liters of a solution that is 50% acid to obtain a solution that is 40% acid?
Practice Questions
1 question
Q1
A chemist has a solution that is 25% acid. How much of this solution must be mixed with 10 liters of a solution that is 50% acid to obtain a solution that is 40% acid?
5 liters
10 liters
15 liters
20 liters
Let x be the liters of 25% solution. The equation is 0.25x + 0.5(10) = 0.4(x + 10). Solving gives x = 15 liters.
Questions & Step-by-step Solutions
1 item
Q
Q: A chemist has a solution that is 25% acid. How much of this solution must be mixed with 10 liters of a solution that is 50% acid to obtain a solution that is 40% acid?
Solution: Let x be the liters of 25% solution. The equation is 0.25x + 0.5(10) = 0.4(x + 10). Solving gives x = 15 liters.
Steps: 13
Step 1: Understand that we have two solutions: one is 25% acid and the other is 50% acid.
Step 2: Let 'x' be the amount of the 25% acid solution we need to find.
Step 3: We know we are mixing this 'x' liters of 25% solution with 10 liters of 50% solution.
Step 4: Calculate the amount of acid in the 25% solution: 0.25 * x.
Step 5: Calculate the amount of acid in the 50% solution: 0.5 * 10 = 5 liters.
Step 6: The total amount of acid in the final mixture will be the sum of the acid from both solutions: 0.25x + 5.
Step 7: The total volume of the final mixture is x + 10 liters.
Step 8: We want the final mixture to be 40% acid, so we set up the equation: (0.25x + 5) / (x + 10) = 0.4.
Step 9: Multiply both sides by (x + 10) to eliminate the fraction: 0.25x + 5 = 0.4(x + 10).
Step 10: Distribute 0.4 on the right side: 0.25x + 5 = 0.4x + 4.
Step 11: Rearrange the equation to isolate 'x': 5 - 4 = 0.4x - 0.25x.
