Understanding the Problem

* Initial Solution: The chemist starts with 'm' grams of saltwater, and a certain percentage of that mass is salt.

* Desired Solution: The chemist wants to end up with a solution that is twice as salty (2 * original salt percentage).

* Goal: Determine how much water needs to be added to achieve this.

Calculations

1. Initial Salt Mass: The initial amount of salt in the solution is (percent/100) * m gm.

2. Desired Salt Mass: To have a solution twice as salty, the final salt mass needs to be 2 * (percent/100) * m gm.

3. Salt Mass Difference: The difference in salt mass between the initial and final solutions is (2 * (percent/100) * m gm) - ((percent/100) * m gm) = (percent/100) * m gm.

4. Water Added: Since the amount of salt remains constant, the difference in salt mass represents the amount of water that needs to be added. Therefore, you need to add (percent/100) * m gm of water.

Example

Let's say the chemist has 100 gm of saltwater that is 5% salty.

* Initial salt mass: (5/100) * 100 gm = 5 gm

* Desired salt mass: 2 * (5/100) * 100 gm = 10 gm

* Salt mass difference: 10 gm - 5 gm = 5 gm

* Water added: 5 gm

Therefore, the chemist needs to add 5 gm of water to make the solution 2 * 5% = 10% salty.