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)
Molarity
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Re: Molarity
Hi!
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!
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!
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Re: Molarity
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
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!
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Re: Molarity
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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