Removing a part of a solution and diluting it?

[Isaac Wen]

Hi! This was a problem from one of the modules we completed before class started. I was just curious about how we should approach problems that involve first dissolving a solution and then diluting only a part of it.

5.00 g of KMnO4 is dissolved in a 150mL flask of water. If 20mL of this solution is removed and placed in a new 2nd 250mL flask and filled with water, what is the concentration of the solution in the 2nd flask? Molas Masses: K (39.10g/mol) Mn (54.94 g/mol), O (16 g/mol)

[Elena Chen 2E]

Hey! Here's how I solved it.

First, find the number of moles that was dissolved in the first beaker (converting grams to moles) and find the concentration of that beaker by dividing the moles you found by the volume they gave. Then, using that concentration, find the number of moles in the amount of solution they asked you to remove (in this case, 0.02 L I believe). This would be concentration times volume to equal moles. Lastly, take that number of moles and divide it by the final volume given (I think it's 0.250 L for this problem), which should give you the final concentration. Remember that the final concentration should be less than the initial one since you're diluting the solution.

I hope this helps!

[Lily Carlson 1K]

I think that you can simplify the approach given by the previous reply even further:

As mentioned, calculate the concentration of the original solution by finding the moles of solute and dividing that by volume. Regardless of how much solution is then removed, the molarity of the remaining solution will be the same (since a proportionate amount of solute and volume is removed!). Then, it's just a simple dilution problem:

If we call the concentration of the original solution C1, we get:
C1*V1=C2*V2
C1*(150-20)=C2*250
Then, solving for C2:
C2=(250/130)*C1=1.92*C1

Then, all that's left is to solve for C1 and substitute it into the equation to find the C2, the resulting concentration!

[Arya Adibi 1K]

For these molarity-dilution type problems, most of the time you will use the relationship: n(initial) = n(final), where n is moles of solute in the solution.
Typically, you will be given or able to derive 3 out of the 4 components: M(initial)V(initial) = M(final)V(final). And you can solve for the value the question asks for.

[Sedge Greenlee]

As Lily said, the proportion of solute and volume removed stays the same regardless of how much is taken out. For these types of problems, it's important to keep in mind these two principles of dilution problems: Unless some amount of solution is taken away, the moles of the given substance stays the same. Conversely, when only using a portion of a diluted substance, the molarity stays the same, not the moles.

[alebenavides]

I had a question: When diluting a solution the moles of solute remain the same correct? I couldn't remember.

[Jaden Kwon 3C]

Yeah when diluting a solution the moles of solute always remain the same. So knowing that molarity = (moles of solute)/(liters of solution) and rearranging to solve for moles of solute gets us moles of solute = (molarity)(liters of solution) or MV. Then we get the equation M1V1 = M2V2 where M1/M2 are initial/final molarities and V1/V2 are initial/final volumes of solution as the moles of solute remain the same before and after the dilution.

[Jaden Kwon 3C]

Yeah when diluting a solution the moles of solute always remain the same. So knowing that molarity = (moles of solute)/(liters of solution) and rearranging to solve for moles of solute gets us moles of solute = (molarity)(liters of solution) or MV. Then we get the equation M1V1 = M2V2 where M1/M2 are initial/final molarities and V1/V2 are initial/final volumes of solution as the moles of solute remain the same before and after the dilution.