4.00L of water was added to 0.400L of an aqueous ethanol solution, giving a final concentration of 0.0400 M. What is the initial concentration of ethanol?

I just need help setting it up pls

Thanks!

Never mind I figured it out!

you diluted it by a factor of 11, with one part original, ten parts water.

Then, if the final concentration is.04M, the initial must have been .44M (11 times the final)

Want to do it with Math?

final solute= initial solute
.04M*4.4L=?*.4L

?= .44M

Sure! To solve this problem, we can use the formula for dilution:

C1V1 = C2V2

Where C1 is the initial concentration of ethanol, V1 is the initial volume of the ethanol solution, C2 is the final concentration of ethanol, and V2 is the final volume of the diluted solution.

Let's assign the variables:
C1 = initial concentration of ethanol (unknown)
V1 = initial volume of the ethanol solution = 0.400 L
C2 = final concentration of ethanol = 0.0400 M
V2 = final volume of the diluted solution = 0.400 L + 4.00 L

We know that the initial volume of the ethanol solution is 0.400 L, and we added 4.00 L of water to it, resulting in a final volume of 0.400 L + 4.00 L = 4.40 L.

Now we can plug in the values into the dilution formula:

C1 * V1 = C2 * V2

C1 * 0.400 L = 0.0400 M * 4.40 L

Now we can solve for C1:

C1 = (0.0400 M * 4.40 L) / 0.400 L

C1 = 0.440 M

Therefore, the initial concentration of ethanol is 0.440 M.