Pre-Module Assessment Q23 and Q25

Moderators: Chem_Mod, Chem_Admin

Posts: 32
Joined: Fri Sep 28, 2018 12:16 am

Pre-Module Assessment Q23 and Q25

Postby 105169446 » Tue Oct 02, 2018 11:13 am

Hello, I am having a difficult time figuring out how to solve the two following questions:

"23. 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? Molar Masses: K (39.10 g/mol), Mn (54.94 g/mol), O (16.00 g/mol)"

"25. A solution is prepared by dissolving 55.1 g of KCl in approximately 75 mL of water and then adding water to a final volume of 125 mL. What is the molarity of KCl(aq) in this solution?"

Could anyone offer an explanation to solve these two problems? Thank you so much in advance.

Nada AbouHaiba 1I
Posts: 77
Joined: Fri Sep 28, 2018 12:28 am
Been upvoted: 1 time

Re: Pre-Module Assessment Q23 and Q25

Postby Nada AbouHaiba 1I » Tue Oct 02, 2018 11:48 am


Okay, to start off #23 is a dilution problem, so we know we want to get it into a M1V1=M2V2 equation. First we're going to solve for M1 andyou're going to want to convert the grams of KMnO4 into mols (so divide by the molar mass).
5.00 g KMnO4/58.04 g/mol = .0316 mol KMnO4
Then you're going to want to find the concentration of this solution by diving the moles by the volume of water, in this case .150 L.
.0316 mol/.150 L= .211 M. (this is our M1)
Then we're told this is transferred to .020 L flask, so that's our intial volume used (V1).
We're then told our final volume will be 250.00 mL (V2).
Finally, we plug everything into the dilution equation and solve for M2.
.211M)(.02 L)=(.25M)x
x=.0169 M

#25 is very similar in concept
So first we're going to want to convert 55.1g KCl into moles by diving it by its molar mass
55.1g/74.55 g/mol = .739 mol.
Then we're going to use the M1V1=M2V2 equation, but in this case remember that M1V1 is the same thing as moles initial, so I'm just going to use that value that we found above as our M1V1 substitute.
We know our final volume(V2) is 125 mL (.125 L)
Lastly, we'll plug everything into our equation and solve for the final concentration
.739 moles = .125L (M2)
M2=5.91 M

Hope this helped! :)

Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

Re: Pre-Module Assessment Q23 and Q25

Postby Sean_Rodriguez_1J » Tue Oct 02, 2018 12:06 pm

23. First for this problem, you will have to find the molar mass of KMnO4. Using the given molar masses, it comes out to be 158.04 g x mol^-1.
By multiplying this by the amount of grams given (5.00), you will get the moles of KMnO4 (5.00 g / 158.04 g x mol(KMnO4)^-1 = 0.0316 moles.) To find the molarity, simply divide by the given amount of Liters (0.150) to get 0.0316/0.150 = 0.211M.

The problem mentions that 0.0200 L of the solution is transferred (V1), so we can now use the equation M1V1 = M2V2 to find the molarity of the final solution (M2). With some division, the above equation can be rewritten as M2 = (M1 x V1)/(V2). Plug the numbers in to find M2.
(0.211 M x 0.020 L) / (0.250L) = 0.0169M

25. This is a similar procedure to #23. First, find the moles of KCl that you have (55.1g / 74.55g x mol(KCl)^-1 = 0.739 mols KCl), then divide by the volume of the first solution (0.739/0.0750 = 9.85M). now we can use the modified equation M2 = (M1 x V1)/(V2) again to find our final molarity, plugging in the numbers accordingly.
(9.85 M x 0.0750 L) / (0.125L) = 5.91M

Return to “Molarity, Solutions, Dilutions”

Who is online

Users browsing this forum: No registered users and 2 guests