posted by MEl .
How would you prepare 2.5 L of a 0.05 M solution of iron (III) chloride using iron (III) chloride hexahydrate
Since 1 mol iron(III)chloride hexahydrate contains 1 mol iron(III) chloride, use the molar mass of the hexahydrate.
M x L = grams/molar mass
Solve for grams. In preparing the solution, BE SURE that you add water to a certain volume and not add a certain volume of water.
Post your work if you get stuck.
Moles iron (III) chloride hexahydrate
Moles = volume x concentration
2.5L (0.05M) = 0.125 moles
Then using number of moles
Mass required = number of moles x (RMM of iron (III) chloride hexahydrate)
0.125moles x 162.2g/mol = 20.27 grams
is this right?
Your answer is ok if you are weighing FeCl3 BUT the problem says to weigh FeCl3*6H2O. So, no, it isn't correct. I thought I said something in my response about using the molar mass of the hexahydrate.
hexahydrate: 270.3 g·mol−1
therefore would it be
0.125 moles x 270.3 g/mol = 33.78 grams?
Yes. Since the question asks how to prepare a solution of xxxxxx, it is important that you understand the difference between making the solution to volume and add a certain volume of water. For example, the directions would be thus:
Weigh 33.78 grams of FeCl3*6H2O, place in a 2.5 L volumetric flask, add some water to dissolve all of the iron compound, THEN add sufficient water to make to the mark in the volumetric flask. Notice that this is different from adding 33.78 g of the iron compound to a flask and adding 2.5 L H2O.
Hello, I have the answer to this question because I did everything empirically, not theoreticaly like all you people. Anyways, first I take a relatively exact amount of iron (III) chloride hexahydrate and by method of titration, I neutralize the iron (III) chloride with a NaOH solution. You people may not understand it so I'll just give you the answer.