concentration of the solution

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Monica Soliman 3F
Posts: 97
Joined: Wed Sep 30, 2020 9:57 pm

concentration of the solution

Postby Monica Soliman 3F » Tue Oct 06, 2020 9:19 pm

On the molarity post assessment, I had a question(attached below) on how to set this up? how do I incorporate molar mass? Thank You
Attachments
Screen Shot 2020-10-06 at 9.05.09 PM.jpeg

Julianna_flores3E
Posts: 91
Joined: Wed Sep 30, 2020 9:52 pm

Re: concentration of the solution

Postby Julianna_flores3E » Tue Oct 06, 2020 9:24 pm

You would need to convert the amount of KMnO4 to moles and the ml into liters. Once you find that, you know the initial molarity and the initial volume then you would use M1V1=M2V2 since you also know the final volume.

Sonel Raj 3I
Posts: 90
Joined: Wed Sep 30, 2020 9:48 pm

Re: concentration of the solution

Postby Sonel Raj 3I » Tue Oct 06, 2020 9:29 pm

You would need to use to molar mass in order to find the number of moles of KMnO4. You would find the molar mass of this molecule using the given values, and use the formula (n=g/MM) to find the number of moles of KMnO4 present. This will help you find the initial molarity and complete the problem.

JSumpter_3I
Posts: 3
Joined: Wed Sep 30, 2020 10:00 pm

Re: concentration of the solution

Postby JSumpter_3I » Tue Oct 06, 2020 9:38 pm

Convert the 5 grams of KMnO4 into moles by dividing it by its molar mass. Then, since you’ll want to be able to the M1V1=M2V2 formula, divide the number of moles by the volume of the original solution (0.15000 L) - this will be M1. Since they’re only using 20.00 mL of the original solution, V1 will be 0.02000 L. V2 will be the 0.25000 L. Plug the three variables in to solve for M2. :)

Navdha Sharma 3J
Posts: 96
Joined: Wed Sep 30, 2020 9:46 pm

Re: concentration of the solution

Postby Navdha Sharma 3J » Tue Oct 06, 2020 9:40 pm

Hi!

I just wrote down the solution. Hope it helps! Also, I used slightly different masses than the question but the answer is the same!
Attachments
new doc 12_1.jpg

alebenavides
Posts: 117
Joined: Wed Sep 30, 2020 9:39 pm

Re: concentration of the solution

Postby alebenavides » Fri Oct 09, 2020 1:46 pm

In order to solve this you need to find the molar mass so in order to do that you would need to convert the grams to moles then divide by the molar mass.

rhettfarmer-3H
Posts: 101
Joined: Wed Sep 30, 2020 9:59 pm

Re: concentration of the solution

Postby rhettfarmer-3H » Fri Oct 09, 2020 3:37 pm

This problem demonstrates a little confusion because we have basically 2 problems that take place which causes 3 volume to be there. We have the first problem that is a simple M=n/v to give us the molarity for the second problem which is a MV=MV. Therefore, we see slight hardships. But first convert to moles from grams using 158.04g/mol and then Find molarity with first volume give with the moles found. Then use molarity with .02 equals new molarity times .250 shown in work.
Attachments
IMG_2758.pdf
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