That is the answer for preparing 1mL.
did you intend on that?
Even though I have no idea what I am doing, I attempted to do the problem:
(5.25 x10^-6/1L)(1L/1000mL)(792.86 g/1 mol)= 4.2 x10^-6 g
Am I correct?
did you intend on that?
M=moles/L
They question tells you that you need 1 L. So, in this case M=moles. The other conversion isnβt required to answer this problem.
To prepare a 5.25 x 10-6 M solution of Brilliant Blue dye, you first need to calculate the number of moles required. The formula weight of the dye is given as 792.86 g/mol.
So, the number of moles required is:
(5.25 x 10-6 mol/L) x (1 L/1000 mL) = 5.25 x 10-9 mol
Now, to find the mass required, you can multiply the number of moles by the molar mass of the dye:
5.25 x 10-9 mol x 792.86 g/mol = 4.15 x 10-6 g
So, to prepare a 5.25 x 10-6 M solution, you would need to add approximately 4.15 x 10-6 grams of Brilliant Blue dye to 1.00 liter of water.
Please note that these calculations assume you have access to pure Brilliant Blue dye, and there are no other factors to consider in the solubility or stability of the dye.
First, let's calculate the number of moles of Brilliant Blue dye required to make a 5.25 x 10^-6 M solution:
Molarity (M) = moles of solute / liters of solution
Given:
Molarity (M) = 5.25 x 10^-6 M
Liters of solution = 1.00 L
Therefore, moles of solute = Molarity (M) x Liters of solution
moles of solute = (5.25 x 10^-6 M) x (1.00 L) = 5.25 x 10^-6 moles
Now, let's use the moles of solute to calculate the grams of Brilliant Blue dye required:
Formula weight = grams / moles
Given:
Formula weight = 792.86 g/mol
moles of solute = 5.25 x 10^-6 moles
Therefore, grams of solute = Formula weight x moles of solute
grams of solute = (792.86 g/mol) x (5.25 x 10^-6 moles) = 4.16 x 10^-3 g
So, you need to add 4.16 x 10^-3 grams of Brilliant Blue dye to 1.00 L of water to prepare a 5.25 x 10^-6 M solution.
Note: When performing calculations like this, it is always good practice to use the correct number of significant figures. In this case, since the molarity was given with 3 significant figures, it is appropriate to report the grams of dye with 3 significant figures as well, giving the answer as 4.16 x 10^-3 g.