how do i figure out the dilutions of 1.5L of a 15.0% (v/v) NaCl/water solution, what would the concentration of NaCl be if you added 3.0L of water
double amount of water, same NaCl, half the concentration
I have a problem with the problem and I also have a problem with the answer. The answer first.
You aren't doubling the volume. Assuming the volumes are additive, and with this concn NaCl they probably are not, the volume is increased from 1.5L to 4.5 L so the new concn is
15 x (4.5/1.5) = 15 x 1/3 = 5%.
Next the problem with the problem. A 15% v/v makes no sense to me with NaCl a solid. How do you measure, for example, 15 mL of NaCl crystals (and what size are they and all of the complications that go with that). If you made a typo and that is 15% w/v you probably need the densities of the two solution; i.e., before and after dilution.
To figure out the dilutions and concentration of a solution, you need to use a formula that relates the volumes and concentrations of the original and diluted solutions.
The formula for dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
In your case:
C1 = 15.0% (v/v)
V1 = 1.5 L
C2 = ?
V2 = 1.5 L + 3.0 L = 4.5 L
First, let's calculate C2 using the formula:
C1V1 = C2V2
(15.0%)(1.5 L) = C2(4.5 L)
C2 = (15.0%)(1.5 L) / 4.5 L
Now, let's calculate C2:
C2 = (0.15)(1.5) / 4.5
C2 = 0.075 mol/L
Therefore, the concentration of NaCl in the final solution, after adding 3.0 L of water to 1.5 L of a 15.0% (v/v) NaCl/water solution, would be 0.075 mol/L.