You have a stock 0.100 M solution of KMnO4 and a series of 100-mL volumetric flasks. What volumes of the stock solution will you have to pipet into the flasks to prepare standards of 1.00, 2.00, 5.00, and 10.0x10^-3M KMnO4 solutions?

I tried using M1V1=M2V2 but the answers didn't make sense

I see what you are saying, but lets start with the first dilution.

You want to prepare a 1 x 10^-3 M solution using 100mL volumetric flasks.

1 x 10^-3 M*(0.1L)= moles of KMnO4 required=1 x 10^-4 moles of KMnO4

1 x 10^-4 moles of KMnO4/stock solution= volume required= 1 mL of stock.

Pipette 1mL of stock into volumetric flask and add solvent until the bottom of the meniscus touches the 100 mL mark.

repeat this setup for the 2.00, 5.00, and 10 x 10 6-3 M solutions.

Oh wow. I just misread the standards the question was asking for. I was trying to make a 1.00M solution rather than a 1x10^-3M. I hope this is what my professor means with the way the question was written.

Let me point out that the dilution formula DOES work.

m1v1 = m2v2
0.1*v = 0.001*100
v = 0.001*100/0.1 = 01/0.1 = 1 mL.:-).

Yes, I used it for my calculations. Thank you.

It happens to all of us.

Good luck!!!!!!!!!!!!!!

In the last line of my response I omitted the decimal point.

It should read as follows:
v = 0.001*100/0.1 = 0.1/0.1 = 1 mL.:-(.

To solve this problem, you can use the dilution equation: C1V1 = C2V2, where C represents concentration and V represents volume.

First, let's determine the initial concentration (C1) and volume (V1) of the stock KMnO4 solution.

Given: C1 = 0.100 M

Now, let's calculate the volume of the stock solution (V1) needed to prepare each of the desired solutions.

1. For the 1.00 M solution (C2 = 1.00 M):
Using the dilution equation, we have:
(0.100 M)(V1) = (1.00 M)(100 mL)
V1 = (1.00 M)(100 mL) / (0.100 M)
V1 = 1000 mL = 1.00 L

Therefore, you would need to pipette 1.00 L (or 1000 mL) of the stock KMnO4 solution into a 100-mL flask to prepare a 1.00 M KMnO4 solution.

2. For the 2.00 M solution (C2 = 2.00 M):
Using the dilution equation, we have:
(0.100 M)(V1) = (2.00 M)(100 mL)
V1 = (2.00 M)(100 mL) / (0.100 M)
V1 = 2000 mL = 2.00 L

Therefore, you would need to pipette 2.00 L (or 2000 mL) of the stock KMnO4 solution into a 100-mL flask to prepare a 2.00 M KMnO4 solution.

3. For the 5.00 M solution (C2 = 5.00 M):
Using the dilution equation, we have:
(0.100 M)(V1) = (5.00 M)(100 mL)
V1 = (5.00 M)(100 mL) / (0.100 M)
V1 = 5000 mL = 5.00 L

Therefore, you would need to pipette 5.00 L (or 5000 mL) of the stock KMnO4 solution into a 100-mL flask to prepare a 5.00 M KMnO4 solution.

4. For the 10.0 x 10^-3 M solution (C2 = 10.0 x 10^-3 M):
Using the dilution equation, we have:
(0.100 M)(V1) = (10.0 x 10^-3 M)(100 mL)

To simplify units, convert the concentration to Molar by moving the decimal places:
(0.100 M)(V1) = (0.0100 M)(100 mL)
V1 = (0.0100 M)(100 mL) / (0.100 M)
V1 = 10 mL

Therefore, you would need to pipette 10 mL of the stock KMnO4 solution into a 100-mL flask to prepare a 10.0 x 10^-3 M KMnO4 solution.

In summary, to prepare the desired KMnO4 solutions, you would need to pipette the following volumes of the stock 0.100 M KMnO4 solution into 100-mL volumetric flasks:
- 1.00 M KMnO4 solution: 1.00 L (1000 mL)
- 2.00 M KMnO4 solution: 2.00 L (2000 mL)
- 5.00 M KMnO4 solution: 5.00 L (5000 mL)
- 10.0 x 10^-3 M KMnO4 solution: 10 mL