Question

Part A 

The forensic technician at a crime scene has just prepared a luminol stock solution by adding 16.0 g of luminol into a total volume of 75.0 mL of H2O. 

What is the molarity of the stock solution of luminol? 

Part B 

Before investigating the scene, the technician must dilute the luminol solution to a concentration of 6.00x10^-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? 

 Part C

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

Answer 1

A)

luminol molar mass = 177.16 g/mol

mass, m = 16.0 g

use:

number of mol,

n = mass/molar mass

=(16.0 g)/(1.772*10^2 g/mol)

= 9.031*10^-2 mol

volume , V = 75 mL

= 7.5*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 9.031*10^-2/7.5*10^-2

= 1.21 M

Answer: 1.21 M

B)

Number of moles = M*V

= 6.00*10^-2 M * 2.00 L

= 0.120 mol

Answer: 0.120 mol

C)

number of mol = 0.120 mol

M = 1.21 M

V = ?

use:

V = number of mol/M

= 0.120 mol / 1.21 M

= 0.0992 L

= 99.2 mL

Answer: 99.2 mL