Help on part "c":

The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol into a total volume of 75.0mL of H2O.

a)What is the molarity of the stock solution of luminol?
answer I got: molarity of luminol solution = 1.43M

b)Before investigating the scene, the technician must dilute the luminol solution to a concentration of 6.00×10−2 M. The diluted solution is then placed in a spray bottle for application on the desired surfaces.

How many moles of luminol are present in 2.00 L of the diluted spray?
answer I got: moles of luminol =
0.120mol

I cannot get the correct answer for "c"...I have tried: 172mL,11.9mL, and 1.19*10^4mL. The only other possibility that I can come up with is: 83.9mL. Would this one be correct?...Or...am I still completely out to lunch???

C)- What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?
Express your answer in milliliters.

duplicate post

Divide pART b TO a

To solve part C, you need to find the volume of the stock solution that contains the same number of moles as the diluted solution.

In part B, you found that there are 0.120 moles of luminol in 2.00 L of the diluted spray.

Now, we can use the formula:

Moles = Molarity x Volume

Rearranging the equation to solve for volume:

Volume = Moles / Molarity

So, in this case, we can substitute the values we know:

Moles = 0.120 mol
Molarity = 1.43 M (from part A)

Volume = 0.120 mol / 1.43 M
≈ 0.0839 L

Since the volume is given in liters, you need to convert it to milliliters.

1 L = 1000 mL

Volume = 0.0839 L x 1000 mL/L
= 83.9 mL

Therefore, the correct answer for part C is 83.9 mL.