IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: Greed444 on October 02, 2010, 07:30:13 am
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on OCT/NOV 2009 paper 52, Q1(d) and (e)
Q1(d)
can someone explain how to fill in the tables?? do we need to guess evrything?
and the Q says that there must be the SAME total final volumes.
Im more confused with the marking scheme saying to use < 6x10^2 mol for monoprotic acid,
how is that related in filling the tables?
another thing is, why [diprotic acid]= 0.5 x [monoprotic acid] ?
Q1(e)
can someone outline a simple answer for this? im VERY weak and lost in this.. if you could
help me step-by-step..
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on OCT/NOV 2009 paper 52, Q1(d) and (e)
Q1(d)
can someone explain how to fill in the tables?? do we need to guess evrything?
and the Q says that there must be the SAME total final volumes.
Im more confused with the marking scheme saying to use < 6x10^2 mol for monoprotic acid,
how is that related in filling the tables?
another thing is, why [diprotic acid]= 0.5 x [monoprotic acid] ?
Q1(e)
can someone outline a simple answer for this? im VERY weak and lost in this.. if you could
help me step-by-step..
Monoprotic acid refers to acids which donate only 1 mole of H+ upon dissociation of 1 mole of acid. Apart from H2SO4 and ethanedioic acid all the other 3 acids are monoprotic.
examples : HCl ---> H+ + Cl-
Only one mole of H+ is formed.
Diprotic acids are those which can donate 2 moles of H+ upon dissociation of 1 mole of acid.
Example : H2SO4 ---> 2H+ + SO42-
To fill the tables now :
Number of moles of NaOH present : 2/1000*30 = 0.06
Since NaOH should be in excess number of moles of monoprotic acids should not exceed 0.06.
So you can state any concentration less than that in your table.
For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H+.
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Oops I did not notice part (e)
Anyway am sure by now you must have realised that ethanedioic acid is a diprotic acid.
Let's say you stated 10.0cm3 of ethanedioic acid with concentration 2.0 mol/dm3
As you can see, number of moles of H+ used is 0.04 moles which is less than 0.06 moles. Therefore it's good.
1. Find the Mr of the salt.
Mr : 2(12) +4(16) + 2 +2(18) = 126
Number of moles = Mass/Mr
You need to produce a solution of concentration 2.0mol/dm3
In 250cm3 you'll have only (2/1000 * 250) = 0.5 moles
0.5 = Mass/126 ----> Mass to be used = 63g
Measure 63g of the salt in the graduated flask and add water upto the 250 mark.
Your solution is done.
NOTE : That is just an example to show you the method. In reality you can't really use 63g. It's way too much. You should calculate it for a reasonable mass of about 5-10g.
Hope you understand. If not, let me know your difficulties and i'll try to elaborate more :)
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Monoprotic acid refers to acids which donate only 1 mole of H+ upon dissociation of 1 mole of acid. Apart from H2SO4 and ethanedioic acid all the other 3 acids are monoprotic.
examples : HCl ---> H+ + Cl-
Only one mole of H+ is formed.
Diprotic acids are those which can donate 2 moles of H+ upon dissociation of 1 mole of acid.
Example : H2SO4 ---> 2H+ + SO42-
To fill the tables now :
Number of moles of NaOH present : 2/1000*30 = 0.06
Since NaOH should be in excess number of moles of monoprotic acids should not exceed 0.06.
So you can state any concentration less than that in your table.
For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H+.
i understand now why we have to use <0.06 moles.
does the [acid] depends mainly only on 1 mole of H+ but not 2 moles of H+?
is that why we have to divide by 2?
how bout for triprotic acid like phosphoric acid?
H3PO4 --> 3H+ + PO4^3- (am i wrong? correct me if im wrong plz ^_^'')
so does that mean we must use less than 0.06/3? since 1mole of the acid provides 3 moles of H+?
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Oops I did not notice part (e)
Anyway am sure by now you must have realised that ethanedioic acid is a diprotic acid.
Let's say you stated 10.0cm3 of ethanedioic acid with concentration 2.0 mol/dm3
As you can see, number of moles of H+ used is 0.04 moles which is less than 0.06 moles. Therefore it's good.
1. Find the Mr of the salt.
Mr : 2(12) +4(16) + 2 +2(18) = 126
Number of moles = Mass/Mr
You need to produce a solution of concentration 2.0mol/dm3
In 250cm3 you'll have only (2/1000 * 250) = 0.5 moles
0.5 = Mass/126 ----> Mass to be used = 63g
Measure 63g of the salt in the graduated flask and add water upto the 250 mark.
Your solution is done.
NOTE : That is just an example to show you the method. In reality you can't really use 63g. It's way too much. You should calculate it for a reasonable mass of about 5-10g.
Hope you understand. If not, let me know your difficulties and i'll try to elaborate more :)
Thanks Bro!!
I finally get it!
so it doesn't matter if i used diffrent volumes and conc and get half of 63g or twice the 63g?
when i use 20cm^3 and 1.0 mol/dm^3, n=0.02 which is <0.06/2,
so to produce 1,0mol/dm3 in 250cm3, 0.25moles needed
and i get mass=31.5g
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i understand now why we have to use <0.06 moles.
does the [acid] depends mainly only on 1 mole of H+ but not 2 moles of H+?
is that why we have to divide by 2?
how bout for triprotic acid like phosphoric acid?
H3PO4 --> 3H+ + PO4^3- (am i wrong? correct me if im wrong plz ^_^'')
so does that mean we must use less than 0.06/3? since 1mole of the acid provides 3 moles of H+?
Yupz........it usually depends on that :)
Basically this is the principle!
I can see that I have been a good tutor :P
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Thanks Bro!!
I finally get it!
so it doesn't matter if i used diffrent volumes and conc and get half of 63g or twice the 63g?
when i use 20cm^3 and 1.0 mol/dm^3, n=0.02 which is <0.06/2,
so to produce 1,0mol/dm3 in 250cm3, 0.25moles needed
and i get mass=31.5g
You're welcome pal :)
You are a very good student.......... :D
Yeah it's alright! But it would be best to calculate a solution using the minimum mass possible ;)
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need helps on May/June 2010 paper52 Q1(c) and (d)
Q1(c)
what do we have to do draw here? i simply drew a solution-filled beaker on top of a stand with a bunsen burner underneath it for boiling. and also a thermometer inserted into the beaker. Is this a good approach or the other way around? and are there better ways to answer this?
Q1(d)
can someone explain why the limited supply of 100g of d.water?
do we have to divide them into beakers
(e.g 10g in one beaker, 20g in the 2nd beaker while 30g and 40g in the 3rd and 4th one)?
what do we have to write to gain a mark of three? :-\
can someone guide me to answer this type of question? i can't really answer this very well.
Thanks :)
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need helps on May/June 2010 paper52 Q1(c) and (d)
Q1(c)
what do we have to do draw here? i simply drew a solution-filled beaker on top of a stand with a bunsen burner underneath it for boiling. and also a thermometer inserted into the beaker. Is this a good approach or the other way around? and are there better ways to answer this?
Q1(d)
can someone explain why the limited supply of 100g of d.water?
do we have to divide them into beakers
(e.g 10g in one beaker, 20g in the 2nd beaker while 30g and 40g in the 3rd and 4th one)?
what do we have to write to gain a mark of three? :-\
can someone guide me to answer this type of question? i can't really answer this very well.
Thanks :)
(c) I guess your diagram should be ok. But you need to add some details.
The question ask to label all the apparatus used and its capacity.
Hence you should state the capacity of the thermometer ( Ex : 100oC) and also that of the beaker used (Ex : 500cm3).
For the measurement of temperature, I would suggest you to make use of two thermometers.
One which is just immersed in the solution, that is its bulb is just in the water. Another one which is just on the surface of the liquid without touching it though. Make use of retort stands and clamps to hold your thermometers in position.
NOTE : Both thermometers should have same range and same sensitivity. This also should be stated in your diagram.
If you add all these, all the three marks should be yours ;)
(d) Yeah, you should divide the water accurately so that you will not need more than 100g. You should be preparing solutions of different concentrations each time using some water so as to keep total mass used constant.
But you should not in any way exceed the use of 100g of deionised water. This is a limitation which you must precisely make sure that you cope with.
Then you need to devise a way such that you obtain at least 5 solutions of different concentration in an appropriate table.
The last mark is for correct calculation of the molality of the different solutions that you need to calculate.
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For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H+.
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Why do we have to divide by 2 and not multiplying??????????? 1mole NaOH reacts with 1mole of sulphuric acid and hence reacts with 2H+
so 0.06moles NaOH reacts with 0.06*2 of H+ Isn't it???????????????? I still didn't understand this point and how can we keep the total final volumes the same in each experiment?????????????????? I really need help urgently cuz the markschemes and the examiner's reports are not very useful............PLZ HELP!!!! thanx in advance :)
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F0r nov 2009 ppr52 I really need an explanation for Q1(f) & Q3(d) the markscheme is not giving an answer.I would really appreciate ur help.
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F0r nov 2009 ppr52 I really need an explanation for Q1(f) & Q3(d) the markscheme is not giving an answer.I would really appreciate ur help.
q1(f)
H = Mc deltaT
you have to take M as the total volume of the solution
H= VT*4.3 * delta T
NaOH is the reagent in excess and HCl is the limiting reagent
therefore you should use the limiting reagent for calculation the heat change for 1 mole
suppose you use 20cm3 of acid of concentration 2.0 mol/dm3
1000cm3 of HCl contains 2.0 mol HCl
20cm3----------- (2.0/1000)*20
=0.04
then 0.04 mole of HCl gives H J of heat (VT*4.3 * delta T)
1 mole----------- (H/0.04)*1 J
Q3(d)
you have to work out the molecular formula let say your mean value for Cu was 2.00
element mass Ar mole /simplest
O 1.00 16.0 0.0625 1.98 =2
Cu 2.00 63.5 0.0315 1
Copper (I) oxide
CuO2
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For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H+.
Why do we have to divide by 2 and not multiplying??????????? 1mole NaOH reacts with 1mole of sulphuric acid and hence reacts with 2H+
so 0.06moles NaOH reacts with 0.06*2 of H+ Isn't it???????????????? I still didn't understand this point and how can we keep the total final volumes the same in each experiment?????????????????? I really need help urgently cuz the markschemes and the examiner's reports are not very useful............PLZ HELP!!!! thanx in advance :)
Nan.........1 mole of sulfuric acid provides 2 moles of H+
2NaOH + H2SO4 ----> Na2SO4 + 2H2O
1 mole of sulfuric acid reacts with 2 moles of NaOH
In your case 0.03 moles of sulfuric acid is required to neutralise 0.006 moles of NaOH because sulfuric acid provides 2H+
Total volume is kept constant so that volume can be used as a measure of the concentration of the solutions. You keep total volume constant by adding water.
If you have any problem, let us know and we'll try to help you as much as you can. :)
Relax dude ;)
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Q3(d)
you have to work out the molecular formula let say your mean value for Cu was 2.00
element mass Ar mole /simplest
O 1.00 16.0 0.0625 1.98 =2
Cu 2.00 63.5 0.0315 1
Copper (I) oxide
CuO2
How did u get an average 2??? my average is 3.983 so the formula of copper oxide is CuO PLZ help..... thanks 4 ur help
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Total volume is kept constant so that volume can be used as a measure of the concentration of the solutions. You keep total volume constant by adding water.
If you have any problem, let us know and we'll try to help you as much as you can. :)
Relax dude ;)
what is the volume of water do we have to use???????
Thanx a lot for ur help..
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Q3(d)
you have to work out the molecular formula let say your mean value for Cu was 2.00
element mass Ar mole /simplest
O 1.00 16.0 0.0625 1.98 =2
Cu 2.00 63.5 0.0315 1
Copper (I) oxide
CuO2
hmm it is a fictitious value :D just replace yours and do the calculation
How did u get an average 2??? my average is 3.983 so the formula of copper oxide is CuO PLZ help..... thanks in advance
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here is your answer
element mass Ar mole /simplest
O 1.00 16.0 0.0625 1.00
Cu 3.983 63.5 0.0627 1.00
answer: CuO
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what is the volume of water do we have to use???????
Thanx a lot for ur help..
This will depend.
Example :
1) First experiment you used 50cm3 of a known concentration of solution.
2) Second experiment you are asked to use 30cm3 of the same solution. Since you need to keep total volume constant(50cm3), add 20cm3 of water ;)
Anytime :)
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here is your answer
element mass Ar mole /simplest
O 1.00 16.0 0.0625 1.00
Cu 3.983 63.5 0.0627 1.00
answer: CuO
which value is a fictitious value??? mine or yours
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This will depend.
Example :
1) First experiment you used 50cm3 of a known concentration of solution.
2) Second experiment you are asked to use 30cm3 of the same solution. Since you need to keep total volume constant(50cm3), add 20cm3 of water ;)
Anytime :)
did we assume that total volume will be 50cm3 in experiment????
Ur explanation is wonderful. You are doing a great job...thank u very much :)
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Ur explanation is wonderful. You are doing a great job...thank u very much :)
You're welcome buddy ;)
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This will depend.
Example :
1) First experiment you used 50cm3 of a known concentration of solution.
Did we assume that the total volume is 50 cm3????
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Did we assume that the total volume is 50 cm3????
Nope that's not an assumption. :D
You'll be given the total volume to be used in the question itself. You'll have to adjust for the other experiments such that you keep it constant for all experiments by adding water.
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which value is a fictitious value??? mine or yours
well the answer is
element mass Ar mole /simplest
O 1.00 16.0 0.0625 1.00
Cu 3.983 63.5 0.0627 1.00
answer: CuO
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Nope that's not an assumption. :D
You'll be given the total volume to be used in the question itself. You'll have to adjust for the other experiments such that you keep it constant for all experiments by adding water.
but the question did not include 50cm3 ::) ::)
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well the answer is
element mass Ar mole /simplest
O 1.00 16.0 0.0625 1.00
Cu 3.983 63.5 0.0627 1.00
answer: CuO
thanx :)
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but the question did not include 50cm3 ::) ::)
Well.........I was not being specific about a particular question!
It was just an example that I stated to help you understand.
Tell me which paper it is and i'll try to help you ;)
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Well.........I was not being specific about a particular question!
It was just an example that I stated to help you understand.
Tell me which paper it is and i'll try to help you ;)
Nov. 2009 ppr52 Q1 (e)
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Nov. 2009 ppr52 Q1 (e)
OK........but that will depend on the concentration you stated in part(d) for ethanedioic acid.
I just finished it now, but I do't have a scanner, so won't be able to upload it right away.
Am sorry I need to rush for now.........i'll be back later to elaborate. Same for your number in p2 Nov 07
Hope you don't mind :)
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ok thanx I will be waiting for ur answer :)
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Nov. 2009 ppr52 Q1 (e)
Let's say the concentration stated above was 2.0 moldm-3
You need to calculate what mass of the salt you would need to form 2.0 moldm-3 solution.
Mr of the solid ethanedioic acid = 2(12) + 4(16) + 2 + 2(18) = 126
Conc in g/dm3 = Mr x Conc in moldm-3
Hence Conc in g/dm3 = 126 x 2.0 = 252
In 1000 dm3 of solution, 252g of this solid will be required.
You need to prepare this solution of concentration 2.0 moldm-3 in a 250cm3 graduated flask.
Hence you will need ( 252/1000 x 250 )= 63g of the solid
Procedure :
1. About 50cm3 of water were placed in a beaker.
2. 63g of ethanedioic acid were then added into the beaker.
3. Stir the mixture until all the solid dissolves.
4. Pour the acid solution in a 250cm3 volumetric flask.
5. Rinse the beaker with a small amount of water and pour the washings into the volumetric flask.
6. Bring up to the mark with distilled water.
Here is your required solution of ethanedioic acid of concentration 2.0 moldm-3 ;)
Hope it helps :)