Expert: Josh - 5/13/2011

Question
http://www.sciencebuddies.org/science-fair-projects/project_ideas/OceanSci_p003.... This is the link to the science experiment. I need help with g.under 3.

Answer
Ashley,

To keep track of how much salt is in each cup, let the initial concentration of the stock solution be X. Without further ado, let's work out X using the given information:
* 1 cup of salt is about 292 g
* 1 qt. of water is 0.946 liters [L].
Because tap water has a density of 1 (or very close to it) 1 mL weighs 1 gram. This occupies 1 cm^3 space at room temperature. [note: 1000 milliliter = 1 liter, so 1 milliliter = 0.001 liter]

So, the concentration of the stock
X = weight of salt / (weight of salt + solvent)
 = 292/(292+946)

Since Cup 1 is filled entirely with the stock solution, its concentration is also X.
For Cup 2, we have mixed equal amount of stock and water. It contains as much salt as the solution in Cup 1. However, with 3/4 cup of stock and 3/4 cup of water, its volume has doubled. So, it has been diluted relative to Cup 1 (which has the same concentration as X, since it was poured from X) by a factor of 2. So, its concentration relative to the stock is X/2.

For Cup 3, since the solution in Cup 2 has half the concentration of the original stock, 3/4 cup is only equivalent to 0.5 * (3/4) cup of the original.

Applying the formula:
new concentration = (volume of stock) / (volume of stock + volume of water)
                 = (0.5 x 3/4) / [(0.5 x 3/4) + (3/4)] * X
                 = X/3

For Cup 4, since the solution in Cup 3 is only one third the strength of the original stock, 3/4 cup is effectively equivalent to (1/3)*(3/4) = (1/4) cup of the original.
Applying the formula:
new concentration = (volume of stock) / (volume of stock + volume of water)
                 = (1/3 x 3/4) / [(1/3 x 3/4) + (3/4)] * X
                 = X/4