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September 19, 2025, 05:13:50 pm

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Martoman

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AAS clarification
« on: June 04, 2010, 08:14:45 pm »
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I would like to be totally clear with this.

A mass of 0.1372g of shellfish is dissolved in 2mL of nitric acid and heated. It is transferred to a 500mL flask, where it is made up to the mark with water. A volume of 1ml of this solution is then further diluted to 250mL in a second flask. The concentration was found to be 0.35 ug / L.

It wants to find the amount of mercury in the shellfish.

Ok so I know that in the 250 there is 0.35 ug/L

1mL of this solution will be the same concentration of the 500mL solution (CORRECT ME HERE IF THIS IS RUBBISH)

so 0.35/1000 ug/ml

here i am now confused. what do i do from here.
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stonecold

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Re: AAS clarification
« Reply #1 on: June 04, 2010, 08:18:23 pm »
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.35/1000= x/250

x=0.0875 ug in 1 mL

0.0875/1=x/500

x= 43.75 ug in the 0.1372 g sample.
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Martoman

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Re: AAS clarification
« Reply #2 on: June 04, 2010, 08:32:08 pm »
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uhhhhh this is where i get confused still.

At the first step you are setting up ratios and saying that:

0.35 micro grams are in 1000ml and then you are finding how many micrograms are in 250ml. how can you say then that this concentration is the same at 1mL as it is in 250?
2009: Math methods: 50, Psychology: 44
2010: chem 47, further 48, Spesh 49 fml seriously and other yr 11 subs.
2011: Holidaying, screw school.
No. Not azn.
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stonecold

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Re: AAS clarification
« Reply #3 on: June 04, 2010, 08:35:32 pm »
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Because the mol never changes.  You can dilute the shit out of it, as long as you use the volume and concentration, you will still get all the mass.

This is why I was doing so shit in AAS for so long.  All the dilutions were confusing me.  THen my teacher told me that the only time which they apply is when you actually take some solution from one thing, and then move it to another AND THEN dilute.  Then you do as I did above. :)

I've never stuffed up a single question since I've been using this method.  It works.  I got the AAS question right on VCAA 2008 today. :)
« Last Edit: June 04, 2010, 08:37:08 pm by stonecold »
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Potter

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Re: AAS clarification
« Reply #4 on: June 04, 2010, 08:35:55 pm »
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.35/1000= x/250

x=0.0875 ug in 1 mL

0.0875/1=x/500

x= 43.75 ug in the 0.1372 g sample.

This is what I got too. The solutions from insight are wrong?
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Martoman

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Re: AAS clarification
« Reply #5 on: June 04, 2010, 08:37:50 pm »
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Nooo what stonecold did was right.

I thought that we were dealing with concentrations!? Like when you do concentrations in aloquits they don't change but inside the system they dooo.... oh sigh...

Edit: Its ok. I get it now. Can ALWAYS think like this as in how much mol there is and use ratios? (this brilliant if so)
« Last Edit: June 04, 2010, 08:50:06 pm by Martoman »
2009: Math methods: 50, Psychology: 44
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stonecold

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Re: AAS clarification
« Reply #6 on: June 04, 2010, 08:53:21 pm »
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Yep, always use the MAXIMUM volume, and there shouldn't be any problems...
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ghadz7

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Re: AAS clarification
« Reply #7 on: June 04, 2010, 09:01:56 pm »
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Shouldn't you find the amount in moles?

2.18 x 10^-7 moles is what I got :s
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Martoman

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Re: AAS clarification
« Reply #8 on: June 04, 2010, 09:08:37 pm »
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Well if i typed it out in full it wanted it in mg. Otherwise yes. find mol as you have.
2009: Math methods: 50, Psychology: 44
2010: chem 47, further 48, Spesh 49 fml seriously and other yr 11 subs.
2011: Holidaying, screw school.
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kenhung123

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Re: AAS clarification
« Reply #9 on: June 06, 2010, 12:28:28 pm »
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We can use concentration or mole ratio right?

Greggler

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Re: AAS clarification
« Reply #10 on: June 06, 2010, 07:17:26 pm »
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yeah same thing was happening with me a while back. sat down with my chem teacher for like half an hour and she pretty much said the same thing to me as your teacher did to you stonecold.

Now i just draw little flow diagrams and whenever an aliquot is taken i write next to it a fraction which refers to how much mole is contained in this aliquot compared to the solution it was taken from.

Makes questions like the last one on vcaa2009 a lot easier

crayolé

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Re: AAS clarification
« Reply #11 on: June 06, 2010, 07:35:28 pm »
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.35/1000= x/250

x=0.0875 ug in 1 mL

0.0875/1=x/500

x= 43.75 ug in the 0.1372 g sample.
Thats what I got Stonecold, but the answers says its this? :/

Solution
Concentration of Hg in the 250.0 mL flask = 0.34 μg L–1

m(Hg) in the 250.0 mL flask = (250.0/1000)x0.34= 0.085 μg
m(Hg) in the 500 mL flask = (500/2.0)x0.085 = 21 μg = 0.021 mg
So, m(Hg) in the shellfish sample = 0.021 mg
« Last Edit: June 06, 2010, 07:38:23 pm by crayola »

stonecold

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Re: AAS clarification
« Reply #12 on: June 06, 2010, 07:46:59 pm »
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Their first step is right.  Second step they've stuffed up.  For some reason, they use 2mL.  But the question Stem says 1mL right?
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crayolé

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Re: AAS clarification
« Reply #13 on: June 06, 2010, 07:52:12 pm »
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Their first step is right.  Second step they've stuffed up.  For some reason, they use 2mL.  But the question Stem says 1mL right?
Their
Okay good, i was fucking pissed last night because I couldn't get the answer no matter how hard i tried lol

Souljette_93

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Re: AAS clarification
« Reply #14 on: June 07, 2010, 02:57:24 pm »
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Their first step is right.  Second step they've stuffed up.  For some reason, they use 2mL.  But the question Stem says 1mL right?


Maybe the 2ml is refers to this;

Quote
;dissolved in 2mL of nitric acid and heated. It is transferred to a 500mL flask, where it is made up to the mark with water. A volume of 1ml of this solution is then further diluted to 250mL in a second flask. The concentration was found to be 0.35 ug / L.


Anyway, doesn't this 2ml play any significance to the procedure, seeming it was diluted to 500ml?


Souljette

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