Luminol has a molecular weight of 177 g/mol. The forensic technician at a crime

Question

Luminol has a molecular weight of 177 g/mol. The forensic technician at a crime scene has just prepared a luminol stock solution by adding 14.0 g of luminol into a total volume of 75.0 mL of H2O. Before investigating the scene, the technician must dilute the luminol solution to a concentration of 6.00×102 M . The diluted solution is then placed in a spray bottle for application on the desired surfaces. What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?

Explanation / Answer

MW = 177 g/mol

m = 14 g of luminol, Vtotal = 75 mL

C final = 6*10^-2 M

initial concentration:

mol = mass/MW = 14/177 = 0.079096 mol

M = mol/V = 0.079096/(75*10^-3) = 1.0546M

now

M1*V1 = M2*V2

V2 = M1/M2*V1 = 1.0546/(6*10^-2)*75

V2 = 1318.25 mL = 1.32 L required

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