Sorry about that
The chemical 5-amino-2,3-dihydro-1,4-phthalazinedione, better known as luminol, is used by forensic scientists in analyzing crime scenes for the presence of washed-away blood. Luminol is so sensitive that it can detect blood that has been diluted 10,000 times. A basic solution of luminol is often sprayed onto surfaces that are suspected of containing minute amounts of blood.
Luminol has a molecular weight of 177g/mol
The forensic technician at a crime scene has just prepared a luminol stock solution by adding 14.0 of luminol into a total volume of 75.0 mL of H2O.
1. What is the molarity of the stock solution of luminol?
ANSWER:molarity of luminol solution =1.05 M
2.Before investigating the scene, the technician must dilute the luminol solution to a concentration of 6.00×10−2 . 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?
ANSWER: moles of luminol =0.120
3.What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?
Answer: needs to be in mL and i cant figure it out. Can someone please help
I think I've answered all of your questions in the first response.
lol i think we are classmate,
answers of
a. 0.753M
b. 0.120 mol
c. dun no
omg!! i kno how to do c now!!
its 159 ml
159
To find the volume of the stock solution (Part A) that would contain the number of moles present in the diluted solution (Part B), we can use the formula:
Volume A = Moles B / Molarity A
Given that the moles of luminol in the diluted solution (Part B) is 0.120 moles, and the molarity of the stock solution (Part A) is 1.05 M, we can substitute these values into the formula:
Volume A = 0.120 moles / 1.05 M
To ensure that the answer is in mL, we need to convert the volume from liters to milliliters. Since 1 L is equal to 1000 mL, we can multiply the volume A by 1000 to get the answer in mL:
Volume A = (0.120 moles / 1.05 M) * 1000 mL/L
Simplifying the equation:
Volume A = 114.29 mL
Therefore, the volume of the stock solution that would contain the number of moles present in the diluted solution is 114.29 mL.