A. The forensic technician at a crime scene has just prepared a luminol stock solution by adding 13.0 g of luminol into a total volume of 75.0 mL of H2O.

-What is the molarity of the stock solution of luminol?

B. Before investigating the scene, the technician must dilute the luminol solution to a concentration of 5.00×10−2 \ 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?

What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?

-

M = mols/L

mols = g/molar mass
Therefore, determine molar mass, divide into grams luminol and divide that by 75 mL (in liters) to obtain molarity = M.

B. M x L = mols
You know M and you know L, calculate mols.
C. Mols = M x L.
You know mols from B, you know molarity (M) from part A, calculate L.
Post your work if you get stuck.

thank you :)

A. To find the molarity of the stock solution, we need to calculate the number of moles of luminol and divide it by the volume of the solution.

The molar mass of luminol is 177.19 g/mol.

Number of moles of luminol = mass of luminol / molar mass of luminol
= 13.0 g / 177.19 g/mol
≈ 0.0734 mol

Molarity (M) = moles of solute / volume of solution in liters
= 0.0734 mol / 0.075 L
= 0.979 M

Therefore, the molarity of the stock solution of luminol is approximately 0.979 M.

B. To find the number of moles of luminol in 2.00 L of the diluted spray, we need to use the molarity and volume of the solution.

Molarity (M) = moles of solute / volume of solution in liters

moles of solute = Molarity × volume of solution in liters
= (5.00×10^(-2) M) × (2.00 L)
= 0.100 mol

Therefore, there are 0.100 moles of luminol present in 2.00 L of the diluted spray.

To find the volume of the stock solution (Part A) that contains the same number of moles as the diluted solution (Part B), we can use the formula:

Volume of stock solution (V1) = moles of stock solution / molarity of stock solution

V1 = 0.100 mol / 0.979 M
≈ 0.102 L
≈ 102 mL

Therefore, approximately 102 mL of the stock solution would contain the same number of moles as the diluted solution.

To find the molarity of a solution, we need to use the equation:

Molarity (M) = moles of solute / volume of solution (in liters)

Let's start with part A:

1. Convert the mass of luminol to moles:
We're given that 13.0 g of luminol is dissolved in 75.0 mL of water. First, we need to convert the mass to moles by using the molar mass of luminol. The molar mass of luminol is 177.18 g/mol.
Moles of luminol = (13.0 g) / (177.18 g/mol)

2. Convert the volume of the solution to liters:
We're given that the volume of the solution is 75.0 mL. To convert it to liters, divide by 1000.
Volume of solution = (75.0 mL) / (1000 mL/L)

Now, we can calculate the molarity of the stock solution (part A):
Molarity = Moles of solute / Volume of solution
Molarity = (13.0 g / 177.18 g/mol) / (75.0 mL / 1000 mL/L)

Moving on to part B:

3. Determine the moles of luminol in the diluted spray:
We're given that the concentration of the diluted spray is 5.00×10^−2 M. This means that for every liter, there are 5.00×10^−2 moles of luminol.
Moles of luminol = (5.00×10^−2 M) * (2.00 L)

Now let's find the volume of the stock solution that would contain the same number of moles as the diluted spray (part B):

4. Rearrange the molarity equation to solve for volume:
Volume of solution (stock) = Moles of solute / Molarity of solution (stock)

Using the moles of luminol from part B and the molarity of the stock solution from part A, we can calculate the volume of the stock solution:
Volume of solution (stock) = (Moles of luminol) / (Molarity from part A)

Remember, we found the moles of luminol and the molarity of the stock solution in the previous calculations.

Simply substitute the values into the equation and calculate.