Let the volume of water used in the mixture be $x$ ml.

Step-by-Step Solution:

1. Amount of salt in the 9% solution:

   • In the 230 ml of 9% salt solution, the amount of salt is: $230 \times 0.09 = 20.7 \, \text{ml of salt}.$

2. Amount of salt in the 8% solution:

   • In the final 270 ml of 8% salt solution, the amount of salt is: $270 \times 0.08 = 21.6 \, \text{ml of salt}.$

3. The amount of salt stays the same after mixing water: Since pure water doesn't contain any salt, the total amount of salt in the final solution comes from the 9% solution.

   Therefore, the salt amount from the 230 ml solution must remain the same in the final 270 ml solution. The only thing that changes is the volume of the solution due to the addition of water.

   We now set up the equation for the total volume: $230 + x = 270.$

4. Solving for $x$: $x = 270 - 230 = 40 \, \text{ml}.$

Thus, 40 ml of water was added to the solution.

---

Would you like more details or have any questions? Here are some related questions to further explore this concept:

1. What would happen if you added salt water instead of pure water?
2. How would the concentration change if 50 ml of water was added instead?
3. What is the minimum amount of salt required to make a 10% solution from 270 ml?
4. How do you calculate the final concentration if the amounts of two solutions differ?
5. How much salt is needed to increase the 8% solution to 12%?

Tip: When mixing solutions, the total amount of the solute (salt in this case) remains constant unless you add or remove it directly.