## Practice Question for Midterm

Lizette Noriega 1H
Posts: 105
Joined: Wed Sep 18, 2019 12:15 am

### Practice Question for Midterm

Hello, could someone explain how to solve for this question? Thanks!

"Potassium permanganate, KMnO4, is an inorganic chemical compound used for cleaning wounds. 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?" - Question 5 on Midterm Practice

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

### Re: Practice Question for Midterm

To solve this problem, we can use the $\textup{M}_1{}\textup{V}_1 = \textup{M}_2{}\textup{V}_2$ equation. To find the original molarity, we have to first convert the 5.00 g of KMnO4 into moles using KMnO4 molar mass ($5.00\textup{g} \textup{KMnO}_4\div158.04\textup{g/mol}= 0.0316 \textup{mol KMnO}_4$). Then divide those moles by 0.250 L to find inital concentration ($0.0316\textup{mol KMnO}_4 \div .150 \textup{L}= 0.211 \textup{mol/L KMnO}_4$).

From the $0.211 \textup{KMnO}_4\textup{mol/L}$, we are using 0.0200 L to create a different molarity. This will be the inital volume. Then the new molarity will contain 0.250 L, so that will be the final volume. The equation will be ($0.211 \textup{M}$ $\textup{KMnO}_4$) $(0.0200\textup{L})$ = $(\textup{M}_2)$$(0.250\textup{L})$

By isolating $\textup{M}_2$, the concentration in the 2nd flask should be $0.0169 \textup{M}$ $\textup{KMnO}_4$.

Btw, I'm just trying out the equation editor, so that's why the numbers and units are like that. Does it actually help or is it better to just keep everything in the same font?

Lizette Noriega 1H
Posts: 105
Joined: Wed Sep 18, 2019 12:15 am

### Re: Practice Question for Midterm

Thank you so much!! That makes more sense. The equation editor is really useful in making sense of the math so thanks for that as well.