# A 280 mL solution is 20% salt. How much water should be added to make the solution 14% salt?

Aug 9, 2015

You need to add 120 mL of water.

#### Explanation:

The key to this problem is to realize that the amount of solute you started with remains unchanged after the solution is diluted.

In essence, the amount of salt you had in the first solution will be found in the final solution as well.

This means that you can determine how much water to add to the first solution by first calculating how much salt you have

"x g salt"/"280 mL solution" * 100 = 20%

This means that

$x = \frac{280 \cdot 20}{100} = \text{56 g salt}$

Let's say that $y$ represents the mass of water that you need to add to the initial solution. This volume of water will impact the percent concetration of the solution by increasing the mass of the solution.

This means that you can write

"56 g salt"/((280 + y)"mL solution") * 100 = 14%

Solve for $y$ to get

$14 \cdot \left(280 + y\right) = 56 \cdot 100$

$3920 + 14 y = 5600$

$14 t = 1680 \implies y = \frac{1680}{14} = \textcolor{g r e e n}{\text{120 mL}}$

So, if you add 120 mL of water to 280 mL of a 20% salt solution, you will get a final solution that is 14% salt.