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The forensic technician at a crime scene has just prepared a luminol stock solution by adding 18.0 g of luminol to H2O creating a solution with a total volume of 75.0 mL. What is the molarity of the stock solution of luminol?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)?

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Answer:

Part A. 1.355 mol/L

Part B. 0.100 mol

Part C. 74.0 mL

Explanation:

Part A.

The molar mass of luminol is 177.16 g/mol, so the number of moles at 18.0 g is:

n = mass/molar mass

n = 18.0/177.16

n = 0.1016 mol

The molarity is the number of moles divided by the volume (0.075 L)

C = 0.1016/0.075

C = 1.355 mol/L

Part B.

The number of moles is the molarity multiplied by the volume, so:

n = 5.00x10⁻² mol/L * 2.00 L

n = 0.100 mol

Part C.

To prepare a solution by dilution, we can use the equation

C1V1 = C2V2

Where C1 is the concentration of the initial (stock) solution, V1 is its volume necessary, C2 is the concentration of the diluted solution, and V2 is its volume.

Thus, C1 = 1.355 M, C2 = 0.05 M, V2 = 2.00 L

1.355V1 = 0.05*2

V1 = 0.074 L

V1 = 74.0 mL

The product of concentration and volume of the stock solution equals the product of concentration and volume of the diluted solution formed

The correct responses are;

Part (A) The molarity of the stock solution is approximately 1.35 M

Part (B) Number of moles of luminol present 2.00 L of the diluted spray is 0.1 moles

Part (C) The volume of the stock solution that would contain the number of moles present in the diluted solution is approximately 73.82 mL

Reason:

Part (A)

Chemical formula of luminol = C₈H₇N₃O₂

Molar mass of luminol = 177.16 g/mol

Number of moles of luminol present, n = [tex]\dfrac{18.0 \ g}{177.16 \ g/mol} = \dfrac{450}{4429} \, moles \approx 1.016 \times 10 ^{-1} \ moles[/tex]

Total volume of solution created, V = 75.0 mL

[tex]Molarity = \dfrac{Number \ of \ moles}{Volume \ in \ liters}[/tex]

Therefore;

[tex]Molarity \ of \ the \ stock \ solution = \dfrac{\dfrac{450}{4429} \ moles}{75.0 \times 10^{-3} \ L} = \dfrac{6000}{4429} \ M \approx 1.35 \, M[/tex]

Part (B)

Required concentration of the luminol = 5.00 × 10⁻² M = 5.00 × 10⁻² moles/liter

Number of moles of luminol present 2.00 L of the diluted spray is therefore;

2.00 L × 5.00 × 10⁻² moles/L = 1.00 × 10⁻¹ moles = 0.1 moles

Number of moles of luminol present 2.00 L of the diluted spray is 0.1 moles

Part (C)

The dilution formula is C₁·V₁ = C₂·V₂

Where;

C₁ = Initial concentration = Concentration of stock solution

V₁ = Initial volume = Volume of stock solution required

C₂ = Final concentration = Concentration of diluted solution

V₂ = Final volume = Volume of diluted solution to be made

[tex]Volume \ of \ stock \ solution \ required = \dfrac{Concentration \times Volume \ of \ diluted \ solution}{Concentration \ of \ stock \ solution }[/tex]

Therefore;

[tex]Stock \ solution \ required = \dfrac{0.1 \, moles}{1.355 \, moles/L } = \dfrac{4429}{60,000} \, L = 7.381\overline 6 \times 10^{-2}\, L[/tex]

Therefore, 7.381[tex]\overline 6[/tex] × 10⁻² L ≈ 73.82 mL  of the stock solution would contain the number of moles present in the diluted solution

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https://brainly.com/question/20709987

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