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A chemist has a solution that is 25% acid. How much of this solution must be mix

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Question: 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?

Options:

  1. 5 liters
  2. 10 liters
  3. 15 liters
  4. 20 liters

Correct Answer: 15 liters

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.

A chemist has a solution that is 25% acid. How much of this solution must be mix

Practice Questions

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?
  1. 5 liters
  2. 10 liters
  3. 15 liters
  4. 20 liters

Questions & Step-by-Step Solutions

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?
  • 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.
  • Step 12: Simplify the equation: 1 = 0.15x.
  • Step 13: Solve for 'x': x = 1 / 0.15 = 15 liters.
  • Concentration and Mixture Problems – This problem tests the ability to set up and solve equations based on the concentrations of different solutions being mixed.
  • Algebraic Manipulation – The question requires the application of algebra to isolate variables and solve for unknowns.
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