Online Module Problem Dilutions

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Elena Maneffa 1E
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Joined: Fri Sep 28, 2018 12:27 am
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Online Module Problem Dilutions

Postby Elena Maneffa 1E » Tue Oct 09, 2018 6:42 pm

I don't understand how to solve this problem. Can somebody help?

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)

Diego Gonzalez 3F
Posts: 32
Joined: Fri Sep 28, 2018 12:26 am

Re: Online Module Problem Dilutions

Postby Diego Gonzalez 3F » Tue Oct 09, 2018 7:10 pm

first, convert the 5.00 grams of KMnO4 to mols. You should get .03164 moles. Then find the molarity by dividing this number by .15 liters, getting .211 M. Then set up a M1V1=M2V2 with M1 being .211, V1 being .02L, and V2 being .25L. Solve for M2 and thats your answer.

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

Re: Online Module Problem Dilutions

Postby Matthew Choi 2H » Sat Oct 20, 2018 1:16 am

First, you need to find the molarity of the original KMnO4 solution. In order to find molarity, you need moles of solute and liters of solution. Use the molar mass of KMnO4 to convert grams to moles, and divide that number by .15000 L. You should end up getting .211 M KMnO4.

Second, you need to use the M1V1 = M2V2 equation to find the final molarity. You have all of the information necessary to finalize the answer.
M1 = .211 M
V1 = .02000 L
M2 = ? (what you are trying to find)
V2 = .25000 L

You're final molarity of the end solution should end up being .0169 M.

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