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?
anwer 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.
I cannot get the correct answer for "c"...I have tried: 172mL,11.9mL, and 1.19*10^4. 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)How many moles of luminol are present in 2.00 L of the diluted spray?
anwer I got: moles of luminol = 0.120mol 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.

Stoichiometry in Chemistry studies about chemical reactions mainly emphasizing quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.

In stoichiometry included :

1. Relative atomic mass
2. Relative molecular mass

is the relative atomic mass of the molecule

3. mole

1 mole is the number of particles contained in a substance with the same number of atoms in 12 gr C-12

1 mole = 6.02.10²³ particles

While the number of moles can also be obtained by dividing the mass (in grams) by the relative mass of the element or the relative mass of the molecule

[tex]\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}[/tex]

Luminol (C₈H₇N₃O₂) is a substance used to detect traces of blood in the scene of a crime because it reacts with iron in the blood

a. a luminol stock solution by adding 19.0g of luminol into a total volume of 75.0mL of H2O.

So the molarity is

1. Luminol mole

- the relative molecular mass of Luminol

= 8. C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams / mol

so the mole:

mol = gram / relative molecular mass

[tex]mole=\frac{19}{177}[/tex]

mole = 0.1073

2. Molarity (M)

M = mole / volume

[tex]M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}[/tex]

M = 1,431

b. luminol concentration in a spray bottle 6.00 × 10⁻² M. So that in 2 L of solution, the number of moles is:

mole = M x volume

mole = 6.10⁻² x 2

mole = 0.12

c. Molarity of the stock solution (Part A) = 1,431 M

the number of moles present in the diluted solution (Part B) = 0.12

So the volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B) is:

volume = mol / M

[tex]volume\:=\:\frac{0.12}{1.431}[/tex]

volume = 0.08386 L = 83.86 ml

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Keywords: mole, volume, molarity, Luminol, the relative molecular mass

dsdrajlin dsdrajlin · Answer 2 · 2021-09-24T20:30:52+02:00

The stock solution prepared by dissolving 19.0 g of luminol in 75.0 mL of water has a molarity of 1.43 M. There are 0.120 moles of luminol in 2.00 L of the 6.00 × 10⁻² M (dilute) solution. The forensic needed 83.9 mL of the concentrated solution to make the dilute one.

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

a) What is the molarity of the stock solution of luminol?

First, we will convert 19.0 g to moles using the molar mass of luminol (177.16 g/mol).

[tex]19.0 g \times \frac{1mol}{177.16g} = 0.107 mol[/tex]

Molarity is equal to the moles of solute divided by the liters of solution.

[tex]M = \frac{0.107 mol}{0.0750L} = 1.43 M[/tex]

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?

The concentration of luminol is 6.00 × 10⁻² M, that is, there are 6.00 × 10⁻² moles of luminol per liter of solution. The moles of luminol in 2.00 L of solution are:

[tex]2.00 L \times \frac{6.00 \times 10^{-2}mol }{L} = 0.120 mol[/tex]

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.

We begin with a solution of concentration 1.43 M (C₁) and prepare 2.00 L (V₂) of a dilute 6.00 × 10⁻² M (C₂) solution. We can calculate the volume of the initial solution (V₁) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\V_1 = \frac{C_2 \times V_2}{C_1} = \frac{6.00 \times 10^{-2} M \times 2.00 L}{1.43 M} = 0.0839 L = 83.9 mL[/tex]

The stock solution prepared by dissolving 19.0 g of luminol in 75.0 mL of water has a molarity of 1.43 M. There are 0.120 moles of luminol in 2.00 L of the 6.00 × 10⁻² M (dilute) solution. The forensic needed 83.9 mL of the concentrated solution to make the dilute one.

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