Exam Suggested Solutions

[bucket]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

[pinchies]

Bingo. Now it makes sense. *self-facepalm*

[Mao]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

[orsel]

"Question 6
a. Give a systematic name for each of the following compounds"

So I guess they'll probably have a fairly lenient marking scheme? Considering that the textbooks are also fairly unhelpful I would be annoyed if they didn't XD.


Thx for the quick response Mao!

[Aba]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.
therefore, to get from diluted 100ml to 25ml aliqout u multiply by (100/25) and then to get from the 25ml aliqout to the 1000ml solution u multiply bu (1000/25)
this is good if u dont use absorbance

[Mao]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

are u sure u wrote the question out corretly?

100% positive.


you ought to stop doubting, aba.
i dont know if you have gotten this wrong [but i think you said its your friend?]

but dilution does not take place in pipettes, and i can assure you that my answer is correct [confirmed by many people, inc teachers]

[bucket]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

But if you're working with the mol, the dilution from 25mL to 100mL will not affect the answer - there is the exact same amount of mol in both solutions, the concentration is just different! :S.
And in that case, the factor would have to be 1000/25, because even though its dissolved, the volume becomes 1L, and only 25mL of that one litre is actually analysed, so therefore there must be 1000/25 times the amount of mol present than what is actually detected. :S

[sally_hates_school]

i think i need to go bury myself in a hole now :uglystupid2:. if all of mao's answers are right i only got 70%  . mao how did you get all of the questions out of the exam? please dont tell me you had enough time to write it on your data book haha. no seriously. or did you get a teachers copy afterwards or something? OR... can you just plain ask the examiners for it?

also, if i wrote 3-chloro-6-methylheptane or pentane or whatever it was... would i only get one mark? was it even out of two marks? (can't remember sorry)

and in anycase... thanks (i think  ) for posting all your answers up here

[Aba]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.
therefore, to get from diluted 100ml to 25ml aliqout u multiply by (100/25) and then to get from the 25ml aliqout to the 1000ml solution u multiply bu (1000/25)
this is good if u dont use absorbance

my friend is here with me, wont give up, he's persuading me

[Mao]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

But if you're working with the mol, the dilution from 25mL to 100mL will not affect the answer - there is the exact same amount of mol in both solutions, the concentration is just different! :S.
And in that case, the factor would have to be 1000/25, because even though its dissolved, the volume becomes 1L, and only 25mL of that one litre is actually analysed, so therefor there must be 1000/25 times the amount of mol present then what is actually detected. :S

dilution is dealing with concentraion, bucket. the reading we took from the calibration curve was actually a reading of concentration rather than amount.

if you were dealing with number of mols [not concentration] (i.e. convert 35mg/L to number of mols in 100mL straight away), then your method will be correct. [and will arrive at the same answer]

[Aba]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

But if you're working with the mol, the dilution from 25mL to 100mL will not affect the answer - there is the exact same amount of mol in both solutions, the concentration is just different! :S.
And in that case, the factor would have to be 1000/25, because even though its dissolved, the volume becomes 1L, and only 25mL of that one litre is actually analysed, so therefor there must be 1000/25 times the amount of mol present then what is actually detected. :S

dilution is dealing with concentraion, bucket. the reading we took from the calibration curve was actually a reading of concentration rather than amount.

if you were dealing with number of mols [not concentration] (i.e. convert 35mg/L to number of mols in 100mL straight away), then your method will be correct. [and will arrive at the same answer]

so my friend is half happy, he may have a slim chance of being right?

[Mao]

i think i need to go bury myself in a hole now :uglystupid2:. if all of mao's answers are right i only got 70%  . mao how did you get all of the questions out of the exam? please dont tell me you had enough time to write it on your data book haha. no seriously. or did you get a teachers copy afterwards or something? OR... can you just plain ask the examiners for it?

i had the opportunity to do both. these are my solutions [in the exam] checked against teacher's solutions after the exam, and except for the "explain" type solutions, the numeric results are pretty all confirmed.

[bucket]

Yes that is what I did! I'm not trying to prove you wrong I'm trying to prove aba wrong =_=.
When I saw the concentration as 35mg/L I converted it to mol straight away, and I'm arguing that the dilution factors play no part in it, so hence the double dilution aba is raving on about is irrelevant!

[Mao]

Yeah look at Q2.
It goes.

13.936g -> Diluted to 1000mL -> 25mL Aliquot -> Diluted to 100mL -> sample goes into AAS.

If you find the mol based on that 35mg/L concentration, you can ignore the dilutions, and all you need to do is multiply the mol by to get the amount of mol in the 1000mL, which you then multiply by the molar mass to find the mass of it in the 13.936g sample.

its not dilute to 1000mL

its basically dissolved to 1000mL[yes, there's a step in between, but that step is not quantitative and does not matter, as it is really just dissolving it]

so we can see, only 1 dilution, but it is the 25mL to 100mL, as the concentration of the 25mL aliquot is the same as the 1L solution

hence the dilution factor is 4, not 1000/25, and there is no other dilution taking place. [full stop]

But if you're working with the mol, the dilution from 25mL to 100mL will not affect the answer - there is the exact same amount of mol in both solutions, the concentration is just different! :S.
And in that case, the factor would have to be 1000/25, because even though its dissolved, the volume becomes 1L, and only 25mL of that one litre is actually analysed, so therefor there must be 1000/25 times the amount of mol present then what is actually detected. :S

dilution is dealing with concentraion, bucket. the reading we took from the calibration curve was actually a reading of concentration rather than amount.

if you were dealing with number of mols [not concentration] (i.e. convert 35mg/L to number of mols in 100mL straight away), then your method will be correct. [and will arrive at the same answer]

so my friend is half happy, he may have a slim chance of being right?

if his answer differs from mine, then no, it'll be wrong. however, methodology is not COMPLETELY wrong, he still might squeeze a point or two from it.

[Aba]

Yes that is what I did! I'm not trying to prove you wrong I'm trying to prove aba wrong =_=.
When I saw the concentration as 35mg/L I converted it to mol straight away, and I'm arguing that the dilution factors play no part in it, so hence the double dilution aba is raving on about is irrelevant!

believe me, the question said further diluted. This is their clue for u. (that rhymes)