Post Module Assessment

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705192887
Posts: 77
Joined: Fri Sep 28, 2018 12:18 am

Post Module Assessment

Postby 705192887 » Thu Oct 11, 2018 1:02 am

So I'm taking the first test tomorrow and I am slightly confused on one of the post-module assessment questions in the Molarity and Dilution section. I was wondering if anyone could just run me through as to how to go about solving this problem? I know to start by finding the initial concentration of KMnO4, but I get confused once the dilution occurs. Thanks!
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Tony Ong 3K
Posts: 63
Joined: Fri Sep 28, 2018 12:23 am

Re: Post Module Assessment

Postby Tony Ong 3K » Thu Oct 11, 2018 1:15 am

First you use molar mass of KMnO4 to find its moles. Convert 5.00 g KMnO4 into moles by dividing by the molar mass, (5.00 g KMnO4)/(158.04 g/mol KMnO4)= 0.0316 mol KMnO4. Then divide by the volume of the solution made to find the concentration, (0.0316 mol KMnO4)/(.15000 L)= 0.211 mol/L KMnO4 solution.
Then use M1V1=M2V2 to solve for M2. the concentration of the new solution knowing that you use 20.00 mL of the original solution and have a total of 250.00 mL of the new solution. (0.211 mol/L KMnO4)*(.02000 L)= (x mol/L)*(0.25000 L). x mol/L= 0.0169 mol/L is the concentration of the new solution.

A De Castro 14B 2H
Posts: 75
Joined: Fri Sep 28, 2018 12:29 am

Re: Post Module Assessment

Postby A De Castro 14B 2H » Thu Oct 11, 2018 1:18 am

Hi! For this equation, you use the Minitial x Vinitial = Mfinal x Vfinal formula. Essentially, what you're asked to calculate is Mfinal. To do that, just rearrange the formula so it becomes Mfinal = (Minitial x Vinitial) / Vfinal.


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