Question

A 120-milliliter mixture of Chemical X and water contained 40 percent Chemical X. A part of the mixture was removed and replaced with an equal quantity of water. If the resulting mixture contained 10 percent Chemical X, what is the volume of the mixture that was removed?

Hint:

A quick way to solve removal and replacement problems is to focus on the substance that is NOT being added (in this case, Chemical X). The formula for the final concentration is: Final Conc. = Initial Conc. \(\times \left(1 - \frac{\text{Volume Replaced}}{\text{Total Volume}}\right)\). Here, \(10% = 40% \times (1 - \frac{V}{120})\). Solving \(\frac{10}{40} = 1 - \frac{V}{120}\) gives \(\frac{1}{4} = 1 - \frac{V}{120}\), so \(\frac{V}{120} = \frac{3}{4}\), which leads to \(V = 90\).

Answer 1

Step 1: Understanding the Concept:
This is a mixture problem involving the removal and replacement of a substance. When a part of the mixture is removed, the amount of the solute (Chemical X) decreases. When water is added, the amount of solute remains unchanged, but the total volume is restored, leading to a dilution and a lower concentration. We need to track the amount of Chemical X throughout the process.
Step 2: Key Formula or Approach:
1. Calculate the initial amount of Chemical X in the mixture.
2. Let \(V\) be the volume of the mixture removed. The amount of Chemical X removed will be \(V\) times the initial concentration.
3. Calculate the amount of Chemical X remaining after the removal.
4. The total volume of the mixture remains the same (120 ml) after replacement with water.
5. Set up an equation where the final amount of Chemical X divided by the total volume equals the final concentration (10%).
6. Solve for \(V\).
Step 3: Detailed Explanation:
The initial total volume of the mixture is 120 ml.
The initial concentration of Chemical X is 40%.
Initial amount of Chemical X = \(40% \text{ of } 120 \text{ ml} = 0.40 \times 120 = 48 \text{ ml}\).
Let \(V\) be the volume (in ml) of the mixture that was removed.
The concentration of Chemical X in the removed portion is also 40%.
Amount of Chemical X removed = \(40% \text{ of } V = 0.40V\).
After removing \(V\) ml of the mixture, the amount of Chemical X remaining is:
Amount of Chemical X remaining = (Initial amount) - (Amount removed) = \(48 - 0.40V\).
An equal quantity (\(V\) ml) of water is then added. This brings the total volume back to 120 ml, but it does not change the amount of Chemical X.
The final mixture has a total volume of 120 ml and contains \((48 - 0.40V)\) ml of Chemical X.
We are told the resulting mixture contained 10 percent Chemical X.
\[ \text{Final Concentration} = \frac{\text{Final Amount of Chemical X}}{\text{Total Volume}} = 10% \] \[ \frac{48 - 0.40V}{120} = \frac{10}{100} = 0.10 \] Now, we solve for \(V\).
\[ 48 - 0.40V = 120 \times 0.10 \] \[ 48 - 0.40V = 12 \] \[ 48 - 12 = 0.40V \] \[ 36 = 0.40V \] \[ V = \frac{36}{0.4} = \frac{360}{4} = 90 \] Step 4: Final Answer:
The volume of the mixture that was removed is 90 ml.