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Lily Benitez 2G
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Joined: Fri Sep 28, 2018 12:29 am


Postby Lily Benitez 2G » Tue Oct 09, 2018 8:58 pm

I have attempted this problem multiple times and keep getting the same answer; which is not one of the answer in the module. I have used the formula M1*V1=M2*V1. For M1 I got .211, V1 is .150L, V2 is .250L, and we are trying to find M2. Did I do a miscalculation?

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)

MaanasO 1A
Posts: 72
Joined: Fri Sep 28, 2018 12:26 am

Re: Molarity

Postby MaanasO 1A » Tue Oct 09, 2018 9:09 pm


For the molar mass of KMnO4, I got (39.10) + (54.94) + 16.00(4) = 158.0 g/mol. This means the moles of KMnO4 is 5.00/158.0 = 0.0316 moles.
The concentration of this initial solution is 0.0316/.150 = 0.211 M.

MiVi = MfVf => Mf = MiVi/Vf = (0.211)*(0.150)/0.250 = 0.127 M. This is final concentration I got. Is this the right answer?

Hope this helps!

Elisa Bass 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

Re: Molarity

Postby Elisa Bass 4L » Tue Oct 09, 2018 9:12 pm

V1 should not be 0.150L because the problem says that only 20.00mL of the 0.211M solution is used in the dilution. Therefore, V1 should be 0.020L.

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Re: Molarity

Postby ariana_apopei1K » Tue Oct 09, 2018 9:18 pm

Hi, I thought of the 150ml solution as a sort of starting point for the person doing the dilution. They would use this to get their first "sample" of 20mL. So try using 20mL as your V1!

Yixiao Hu 3C
Posts: 45
Joined: Fri Sep 28, 2018 12:19 am

Re: Molarity

Postby Yixiao Hu 3C » Sat Oct 13, 2018 3:10 pm

you shouldn't use m1v1=m2v1 this time. the equation you should use is that the mole of kmno4 is the same. so you should calculate the moles of kmno4 in the 20ml solution first.

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