Module Question Help

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Jedrick Zablan 3L
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Joined: Sat Aug 17, 2019 12:17 am

Module Question Help

Postby Jedrick Zablan 3L » Thu Oct 10, 2019 7:55 pm

Can someone help me with this question from the module pls? 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).

Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: Module Question Help

Postby nicolely2F » Thu Oct 10, 2019 8:10 pm

We start by calculating the molar mass of KMnO4, which is 158 g/mol. If 5 grams are mixed into 150 mL, this solution will 3.17*10^-2 moles of KMnO4, which means its concentration is 0.21 mol/L. Therefore, 20mL of such solution will have 4.2*10^-3 moles of KMnO4. If we take these 20 mL and mix them into another 230 mL of water, the new solution will have a concentration of 4.2*10^-3 moles / 250 mL = 1.68*10^-2 M. Hope this helps!
Edit: remember that M is the unit for concentration of solution and is the same thing as saying "mol/L".

Alexa Mugol 3I
Posts: 54
Joined: Sat Aug 17, 2019 12:17 am

Re: Module Question Help

Postby Alexa Mugol 3I » Thu Oct 10, 2019 8:30 pm

The question seemed a little confusing at first for me too, especially with the extra numbers. But basically, you first find the concentration of the the first solution (5g in 150 mL) by converting to moles and then dividing by the volume (in L).
The solution has the same concentration no matter what volume of it you take because when you pour a certain amount out, the moles solute and liters solvent both decrease but stay the same ratio the whole time. So, 150 mL of the solution still has the same concentration as 20 mL of it. It will be diluted when you pour it in the new flask of 250 mL.
In this, you'll use M1V1=M2V2. M1 will be the concentration you just calculated, V1 will be 20 mL, and V2 will be 250 mL. Solve for M2.

Hope this helps!

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