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

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 6.00×10−2 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)?

Expert Solution

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

queen_honey_blossom answered 2 years ago

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