Audio Visual Post Assessment

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Bailey Giovanoli 1L
Posts: 110
Joined: Wed Sep 30, 2020 9:50 pm

Audio Visual Post Assessment

Postby Bailey Giovanoli 1L » Sat Oct 24, 2020 1:47 pm

On number 25 of the audio visual assessment for molar it’s and dilution you are asked:

25. 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)

I know thins involves the m(initial)v(initial)=m(final)v(final).
To start I convert the 5.00 g of KMnO4 to moles and get 0.0316375601 moles.
I then use this to find the molarity of the 150.00 mL solution.
M= 0.0316375601/.150=0.210917673 mol/L
Next I use the formal I stated first to find the molarity of a solution with .250 L
M(.250)= (0.0210917673 x .150)/0.250

Am I missing a step because after the calculations I do not get an answer that is listed?
I’ve tried converting it to 20 mL and then to 250 mL, but nothing seems to work for me.
What am I doing wrong? This is the only problem I can’t seem to get.

Olivia Smith 2E
Posts: 87
Joined: Wed Sep 30, 2020 9:43 pm

Re: Audio Visual Post Assessment

Postby Olivia Smith 2E » Sat Oct 24, 2020 2:07 pm

You may have had a simple calculation error. Starting with molar mass. As I got the total number of moles to be .0352 of KMnO4. Which will mess up your molarité calculations and the whole bunch. I got 0.2346M! Making the final Morality .01877 M in the new container

Also in looking at it again it seems that you have used the M(initial) V(initial) = M (final) V(final) incorrectly

The initial volume is going to .02L final volume will be .25L
Last edited by Olivia Smith 2E on Sat Oct 24, 2020 2:16 pm, edited 2 times in total.

Hannah_Kim_1I
Posts: 97
Joined: Wed Sep 30, 2020 9:56 pm
Been upvoted: 1 time

Re: Audio Visual Post Assessment

Postby Hannah_Kim_1I » Sat Oct 24, 2020 2:11 pm

So it seems like you did everything right until the last equation m(initial)v(initial)=m(final)v(final). What you want to do is plug in the molarity of the 150 mL solution, 0.210917673 mol/L, and the volume removed from that solution, 20 mL, into m(initial)v(initial) respectively. Since the question is asking for the final concentration and gives you the the final volume, 250 mL, the other side of the equation should be m(.250L). Then just solve for m(final).

(0.21 mol/L)*(.020 L) = (m final)*(.250 L)
((0.21 mol/L)*(.020 L))/(.250 L) = (m final)
0.017 mol/L = m final

Bailey Giovanoli 1L
Posts: 110
Joined: Wed Sep 30, 2020 9:50 pm

Re: Audio Visual Post Assessment

Postby Bailey Giovanoli 1L » Sat Oct 24, 2020 2:15 pm

Okay that makes so much more sense. I tried it that way, but it didn’t work. I think it’s because I was using unrounded numbers the whole time. Thank you:)


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