Module Assessment Question

[Angela Ueng 1E]

I wasn't sure how to solve the follow question from the post-module assessment:

5.00 g of KMnO4 dissolved in 150 mL flask of water. If 20 mL of this solution is removed and placed in a new second 250 mL flask and filled with water, what is the concentration of the solution in the second flask? (Molar masses: K 39.10 g/mol; Mn 54.94 g/mol; O 16 g/mol)

I started off by calculating the molar mass of KMnO4 (158.04 g/mol) and converting that into moles (0.0316 mol). I then calculated the molarity of the first solution: M = 0.0316 mol / 0.150 L = 0.211 mol/L. To calculate the molarity of the second flask, do I use the same moles of solute? Also, the volume of the second flask is still 250 mL, not 270 mL, right?

[Danielle Sumilang - 1F]

Hi Angela!

To find the concentration of the solution in the second flask, you should use the equation M (initial) x V(initial) = M (final) x V (final) and try to solve for M (final). Plug in the molarity of the solution in the first flask, 0.211 m/L, as your M (initial), 0.2 L as your V (initial), since you are removing 0.2 L from the first flask, and plug in 0.250 L as your V(final). Then you should be able to find the final molarity.

[Angela Ueng 1E]

Thanks so much for the clarification and help! :)

[Isobel Tweedt 1E]

Also just a note on working with molarity and liquids: when working with liquids the total solution volume changes (make sure to pay attention to the units your liquid is measured in). In this problem we are dealing with a 250ml flask and a solid solute (as seen with units grams) so the total volume isn't changed.