Water Added to Solution Placed in a New Flask

[Meigan Wu 2E]

5.00 g of KMnO4 is dissolved in a 150.00 mL flask of water. If 20.00 mL of this solution is removed and placed in a new 2nd 250.00 mL flask and filled with water, what is the concentration of the solution in the 2nd flask? Molar Masses: K (39.10 g/mol), Mn (54.94 g/mol), O (16.00 g/mol)

How do you solve this problem since only 20.00 mL is removed and then diluted to 250mL?

[Chris Freking 2G]

Since only 20 mL of the first solution is diluted, you should first find the molarity of the first solution and using that molarity and 20 mL volume as your M1V1 later on.

Start by finding the molarity of the original solution. Convert the 5.00g of KMnO₄ into moles by dividing it by the molar mass.
(5.00g KMnO4)/(158.04g/mol KMnO4) = 0.0316 mol KMnO4.

Divide that by the volume of solution to get the molarity.
(0.0316 mol KMnO4)/(0.15000 L) = 0.211 mol/L KMnO4

Then use M1V1 = M2V2 to solve for M2.
(0.211 mol/L KMnO4)*(0.02000 L) = M2 * (0.25000 L)
M2 = 0.0169 mol/L

[505095972]

You can use the equation Vinitial*M initial = Vfinal*M final to solve for M final

You find M initial by converting the 5.00 g KMnO4 to moles per liter using the known 150 mL volume and stoichiometry.

Then you put that concentration and volume into the equation and solve for M final.

[Ariel Cheng 2I]

I think here we need to use the M1V1 = M2V2 formula.
I would first find the molarity of the original solution by converting 5.00 grams of KMnO4 into moles for the moles of solute and then divide it by the volume of the solution.
(0.0316 mol/0.1500 L = 0.211 mol/L)
Now you have the initial molarity (0.211 mol/L), the initial volume (0.020 L), and the final volume (0.250 L).
Plug these into the formula:
(0.211 mol/L) * (0.020 L) = Mfinal * (0.250 L)
Solve for Mfinal and you should get 0.0169 mol/L as the final concentration