Molarity and Dilution of a Solution Post Module Assessment Question 25

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Alexa_Henrie_1I
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Joined: Fri Sep 29, 2017 7:03 am

Molarity and Dilution of a Solution Post Module Assessment Question 25

Postby Alexa_Henrie_1I » Tue Oct 02, 2018 3:50 pm

I am having a hard time solving question 25 on the Molarity and Dilution post assessment.
The question is: 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)

could someone walk me through the steps of this problem? I don't know what I am missing. Thank you

Layal Suboh 1I
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Joined: Fri Sep 28, 2018 12:23 am

Re: Molarity and Dilution of a Solution Post Module Assessment Question 25

Postby Layal Suboh 1I » Tue Oct 02, 2018 5:12 pm

First, you would need to convert grams of KMnO4 to moles:

5g KMnO4/158.04 g/mol= 0.0316 mol

Then, you would need to calculate the initial molarity of this solution by dividing by the initial volume:

0.0316 mol KMn04/ 0.15 L= 0.2109 M KMnO4

Now, you use the MinitialVinitial=MfinalVfinal equation to find the final volume:

(0.2109 M KMnO4)(0.02 L)= (Mfinal)(0.25 L)
Mfinal= (0.2109 M KMnO4 * 0.02 L)/(0.25 L)
Mfinal= 0.0619 M KMnO4

Hope this helps!

Chloe Thorpe 1J
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Joined: Fri Sep 28, 2018 12:16 am

Re: Molarity and Dilution of a Solution Post Module Assessment Question 25

Postby Chloe Thorpe 1J » Tue Oct 02, 2018 7:06 pm

This will be quite similar to Layal's post, but it may help to see it in a slightly different way!

First, I found molar mass of KMnO4:
(molar mass K) + (molar mass Mn) + 4(molar mass O) = 39.10g/mol + 54.94g/mol + 4(16.00g/mol) = 158.04 g/mol

Then, use KMnO4's molar mass to find how many moles of KMnO4 there are:
5.00g/158.0 g/mol = 0.0316mol

I calculated the initial Molarity [I refer to initial Molarity as Mi, and final Molarity as Mf):
M = n/V = 0.0316mol/.150L = 0.211M

Now that we know these values, I like to list the known and unknown values so far.
Mi = 0.211 mol/L
Vi = 0.020 L
Mf = ?
Vf = 0.250 L

Then, I use these values along with the formula, MiVi = MfVf, to solve for the unknown.
MiVi = MfVf
Mf = (MiVi)/Vf = ((0.211 mol/L)(0.020 L))/(0.250 L) = 0.0169 mol/L = 1.69 * 10^-2 mol/L

So the concentration of the solution in the second flask is 1.69 * 10^-2 mol/L.
Hope this helped you!

Alexa_Henrie_1I
Posts: 61
Joined: Fri Sep 29, 2017 7:03 am

Re: Molarity and Dilution of a Solution Post Module Assessment Question 25

Postby Alexa_Henrie_1I » Wed Oct 03, 2018 1:06 pm

Thank you both so much! I just wasn't using .02 L as the initial volume in the equation. Seeing it step by step helped a lot!

melissa_dis4K
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Joined: Fri Sep 28, 2018 12:28 am

Re: Molarity and Dilution of a Solution Post Module Assessment Question 25

Postby melissa_dis4K » Thu Oct 04, 2018 4:08 pm

Can someone please explain why the initial volume is 0.02 L? I thought it was 0.15 L? Thank you!

Chloe Thorpe 1J
Posts: 77
Joined: Fri Sep 28, 2018 12:16 am

Re: Molarity and Dilution of a Solution Post Module Assessment Question 25

Postby Chloe Thorpe 1J » Fri Oct 05, 2018 1:05 pm

@Melissa:

The initial volume is 0.020L because only 0.020L of the original solution is diluted. The 0.150L is there so that you can find the initial concentration of the solution. Hope that makes sense!


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