Examkin

Question

Difficulty: HardMixture and Concentration

A container holds a solution of acid and water that is 50%50\% acid by volume. First, 2020 liters of the solution are removed and replaced with 2020 liters of pure water, resulting in a solution that is 40%40\% acid by volume. Next, 2525 liters of this 40%40\% solution are removed and replaced with 2525 liters of pure acid. What is the percentage of acid, by volume, in the final solution?

  1. A
    47.5%47.5\%
  2. B
    52.5%52.5\%
  3. 55%55\%Answer
  4. D
    65%65\%
  5. E
    70%70\%

Answer

The final concentration of acid by volume is 55%55\%.
The solution requiring 55%55\% concentration is correct. From the first replacement step, we establish that the total container volume is 100100 liters (since removing 2020 liters of 50%50\% solution leaves 0.50V−100.50V - 10 liters of acid, and 0.50V−10V=0.40\frac{0.50V - 10}{V} = 0.40 yields V=100V = 100). Before the second replacement, the container holds 4040 liters of acid. Removing 2525 liters of this 40%40\% solution removes 1010 liters of acid, leaving 3030 liters of acid in 7575 liters of solution. Adding 2525 liters of pure acid increases the acid amount to 5555 liters in a total volume of 100100 liters, producing a final concentration of 55%55\%.

Step-by-Step Solution

1
Determine the total volume of the container (VV) using the first replacement step.
Initial acid volume is 0.50V0.50V. Removing 2020 liters of solution removes 0.50×20=100.50 \times 20 = 10 liters of acid. Replacing with 2020 liters of pure water restores total volume to VV with an acid content of 0.50V−100.50V - 10. Setting up the new concentration equation: 0.50V−10V=0.40  ⟹  0.50V−10=0.40V  ⟹  0.10V=10  ⟹  V=100\frac{0.50V - 10}{V} = 0.40 \implies 0.50V - 10 = 0.40V \implies 0.10V = 10 \implies V = 100 liters.
Finding the fixed total volume of the container is required to calculate the exact amounts of acid removed and added in subsequent steps.
2
Calculate the amount of acid remaining after removing 2525 liters of the 40%40\% solution.
Before the second step, the container has 100100 liters of 40%40\% acid solution, which contains 0.40×100=400.40 \times 100 = 40 liters of acid. Removing 2525 liters of this solution removes 0.40×25=100.40 \times 25 = 10 liters of acid. Acid remaining in the container =40−10=30= 40 - 10 = 30 liters.
When solution is removed, both solute (acid) and solvent (water) are removed proportional to their concentration.
3
Calculate the final acid concentration after adding 2525 liters of pure acid.
Adding 2525 liters of pure (100%100\%) acid adds exactly 2525 liters of acid. Total final acid =30+25=55= 30 + 25 = 55 liters. Total final volume =(100−25)+25=100= (100 - 25) + 25 = 100 liters. Final concentration =55100×100%=55%= \frac{55}{100} \times 100\% = 55\%.
The new concentration is the total volume of acid divided by the total volume of the solution.

Key Concept

Multi-stage removal and replacement in mixture problems
Estimated Time:2m 0s
Rate this question

Topics

Mixture and Concentration

More questions from this topic