Iota Posted October 11, 2012 Posted October 11, 2012 I have 42.1167g of Na2CO3. DIVIDED BY RRM (105.99) = 42.1167/105.99 = 0.397mol. Now, I want to find out how many moles there are in 250cm3 when the powder is dissoved in water. Would that be: 0.397*(250/1000)= 0.0993mol? Then, therefore, would (0.0993/250) *1000 get to moldm-3, molarity? I've repeatedly confused myself with something I was able to do a few weeks ago... and a couple of years before that.
Fuzzwood Posted October 11, 2012 Posted October 11, 2012 Why would the amount of moles change when you dissolve it? Is it any different to dissolve the 42.1167 grams?
Iota Posted October 11, 2012 Author Posted October 11, 2012 I'm not sure. I think I've fallen for an old trick. The question paper first says calculate the moles... as I did simply using g/gmol-1(of Na2CO3); then there's a carry-along question that goes on to ask, how many moles are in 250cm3. The same, then? They way it's laid out suggests the calculation is progressing, so I assumed some sort of change to the mol. Consecutively it asks what the molarity is.
Axioms Posted October 11, 2012 Posted October 11, 2012 The question doesnt make sense. If you have a certain mass it wont change unless you getting a mol percentage. Can you type the questions they give exactly then maybe I can help quickly. Im still waiting to see if anyone replies to my HW post >.<
Iota Posted October 11, 2012 Author Posted October 11, 2012 From the weight of Na2CO3 taken, calculate the number of moles of Na2CO3. (i) Molar mass of Na2CO3 = 105.99gmol-1 (ii) Number of moles Na2CO3= ? (given the mass of it I weighed 42.1167g) (iii) number of moles in 250cm3 = ? (iv) number of moles if you had a 1000cm3 of this solution= ? (Concentration)
Axioms Posted October 11, 2012 Posted October 11, 2012 Question 1 and 2 differ from the rest in my opinion. From question 3 Do they give you the density of Na2CO3? You can calculate how many moles it would be from mass = density * volume. You can then calculate how many mols are in that mixture. The concentration from this amount can then be calculated by taking the (calculated mols/ 750cm^3 H2O). This will give a unit of mols NaCO3/cm^3 H2O. If you need it per solution you can just manipulate the denominator. (calculated NaCO3 mols / 1000cm^3 sol). I dont know what units you want but I'll leave that to you. Concentration = kmol/m^3(sol) or kg/m^3(sol) in my experience unless you working with small values. The only relation I can see between the questions is if you have to calculate the density from what you had initially measured. i.e you take the amount you weighed and found what volume it occupies. This would be the bulk density and there are ways around it but then again I dont know if you got these values experimentally. This is how I would do it =/
hypervalent_iodine Posted October 12, 2012 Posted October 12, 2012 The number of moles from (ii) to (iii) isn't changing if you put the entire 42.1167 g into the solution. Molarity is expressed as moles / L, so question (iv) should be fairly straight forward given your answer to (iii).
Axioms Posted October 12, 2012 Posted October 12, 2012 The moles would change. The OP doesnt say a solution of 250cm^3 of Na2CO3. If it is a solution then the moles will stay the same and then in iv) You could work out as the concentration in 250cm^3 = Xmol Na2CO3/ 1000cm^3. hmm I see in the first post you did say it is in water for the 250cm^3. Lol I just went on what you had said in the second set of questions. hypervalent_iodine Empress of EverythingThe number of moles from (ii) to (iii) isn't changing if you put the entire 42.1167 g into the solution. Molarity is expressed as moles / L, so question (iv) should be fairly straight forward given your answer to (iii). This is a good description.
Iota Posted October 12, 2012 Author Posted October 12, 2012 This is a good description. Indeed it is. Cheers for the help, both.
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