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G.5 Part A) Molarity

Posted: Fri Apr 06, 2018 7:01 pm
by Linda Arroyo 3J
So for part A, I tried to find the molarity but I kept getting 0.004979 and I do not know what I am doing wrong, On the solutions its something else, can someone please help me :)

Re: G.5 Part A) Molarity  [ENDORSED]

Posted: Sat Apr 07, 2018 9:10 am
by Chem_Mod
Sure, we can help! Just post the question and your work, and we'll take a look! Remember to always post the question at least so we can reference it.

Re: G.5 Part A) Molarity

Posted: Sat Apr 07, 2018 2:48 pm
by Bree Perkins 1E
I also don't understand this problem. I tried looking at the solution guide to see how it went about solving it, but I am a bit confused on what the steps are that the solution is doing. Would someone be able to explain the steps to this problem?

Re: G.5 Part A) Molarity

Posted: Sat Apr 07, 2018 3:34 pm
by Kuldeep Gill 1H
Okay so the way I understood this is that we start with 2.111g of a solid and we know we have a flask with a volume of 250 mL part a asks what volume of solution should be transferred to get 2.15 mmol of Na+. So to start out with convert 2.111g to moles by dividing by the molar mass of sodium carbonate which is 105.99. dividing will give you moles of Na2CO3 so 2.111g/105.99g/L=.0199 moles now we find the concentration of the moles in this volume so convert mL to L by dividing by 1000 and you'll get .250L. Now just do the molarity equation M=.0199/.250L and you find that the molarity is .0796 mol/L. Now look at the ratios you need 1 Na2CO3 to get 2 Na+ so the ration is 1:2 now from here multiply the molarity of Na2CO3 by 2 to get the molarity of Na+ you will then do the molarity equation again but this time you know the molarity and you know the millimoles... solve for volume and you should get .0135 L... hope that helps!!!

Re: G.5 Part A) Molarity

Posted: Sun Apr 08, 2018 10:59 am
by AnnaYan_1l
I understand/agree with how Kuldeep got the answer to part A! I just have a couple quick questions...How does the portion of the question that says, "Some of this solution was transferred to a buret" relate to the question? Also, this might be a dumb question but how would you know the molecular formula for sodium carbonate without looking it up? Are there a few molecular formulas (besides the obvious H2O, etc.) that we have memorized? Thanks!

Re: G.5 Part A) Molarity

Posted: Sun Apr 08, 2018 11:40 am
by Chem_Mod
How does the portion of the question that says, "Some of this solution was transferred to a buret" relate to the question

This is more related to how one would probably dispense the solution experimentally. You should note that concentration despite the transfer is unaffected.


but how would you know the molecular formula for sodium carbonate without looking it up?

To know this, you need to know the polyatomic ions. Carbonate is a very common polyatomic ion (see sections B-D of fundamentals) that has a 2- charge. So, 2 Na+ is needed to make the compound neutral.

Re: G.5 Part A) Molarity

Posted: Sun Apr 08, 2018 11:49 pm
by Julia Shapero 1E
So I have a more general question about this type of question: For questions that are asking for the volume, how can I differentiate between questions that are looking to have you use the molarity = moles/volume formula (like this one) and those that are looking to have you use the m(initial)v(initial) = m(final)v(final) formula? (sorry for the excessive use of the word question)

Re: G.5 Part A) Molarity

Posted: Mon Apr 09, 2018 11:56 am
by alexagreco1A
I am still confused about how to do part A, as the solution in the book uses 1 mol Na2CO2 instead of Na2CO3. Where does this come from?

Re: G.5 Part A) Molarity

Posted: Tue Apr 10, 2018 9:04 pm
by Sarah Brecher 1I
Also, why wouldn't you use the volume of 250 mL given in the first sentence of the question and use the equation M(initial)V(initial)=M(final)V(final)?

Re: G.5 Part A) Molarity

Posted: Thu Apr 12, 2018 8:18 am
by Kuldeep Gill 1H
The part that says it was transferred relates to us knowing that a new volume may have to be obtained which is what the question asks

After finding the concentration when transferring it into a new container the concentration should not change so that's why you set the concentration equal to x/.5L

Not sure if that answers you question!.....