To solve this problem, we need to dilute the chemical solution from 95% purity to 50% purity by adding water. Here's how to approach it:

Step-by-step solution:

1. Initial Amount of Pure Chemical X:

   • You have 1 liter of chemical X, which is 95% pure.
   • This means the amount of pure chemical X in the solution is: $\text{Amount of pure chemical} = 1 \times 0.95 = 0.95 \text{ liters}$

2. Final Solution:

   • After adding water, the final solution should be 50% pure, meaning 50% of the total volume will be pure chemical X.
   • Let $V$ be the final total volume (in liters) of the solution after adding water.
   • Since the amount of pure chemical X remains constant at 0.95 liters, the final solution must satisfy: $\frac{0.95}{V} = 0.50$

3. Solve for $V$: $V = \frac{0.95}{0.50} = 1.9 \text{ liters}$

4. Amount of Water to Add:

   • The total volume of the solution after adding water will be 1.9 liters.
   • Since you started with 1 liter of chemical X solution, the amount of water to add is: $\text{Water to add} = 1.9 - 1 = 0.9 \text{ liters}$

Final Answer:

You need to add 0.9 liters of water to dilute the chemical X to a 50% solution.

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Do you need further details or have any questions? Here are some related questions to expand on the topic:

1. How would the process change if the starting solution were only 75% pure?
2. What if you wanted to dilute the solution to 25% purity instead of 50%?
3. How would you handle this problem if the volume of the initial solution was different?
4. Could you calculate the concentration if a different amount of water was added?
5. What is the minimum purity achievable if you keep adding water indefinitely?

Tip: When diluting solutions, the amount of the solute (pure chemical) remains constant, only the total volume changes!