Post Module Assessment

[705192887]

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!

[Tony Ong 3K]

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]

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