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Author Topic: Working out the concentration of a stock solution..  (Read 12490 times)

Offline chloeukc

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I have an online test and I'm really stuck on one of the questions. I dont necessarily need the answer, just being told how to do it would be a great help!!!!!

Calcium in a sample solution was determined by atomic absorption spectrophotometry. A stock solution of calcium was prepared by dissolving 3.668 g CaCl2.2H2O in water and diluting to 1000 cm3. What is the concentration of calcium in this stock solution in ppm?


 

Offline eric l

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Working out the concentration of a stock solution..
« Reply #1 on: 30/10/2007 08:35:16 »
What precision do you want ?  You made the solution by diluting to a volume, and want the concentration by weight.
You could have had that volume in "Mohr litres", which would have excluded doubts, because that obsolete unit was defined as the volume of 1 kg of pure water at the given temperature.  But les us accept the density of water - and of our solution - as 1 g/cm3.
Now we look for the molecular weight of our compound.  Wikipedia (http://en.wikipedia.org/wiki/Calcium_chloride) gives us 147.02.
So the concentration of the solution is (3.668 / 147.02) moles/l.
Each molecule contains 1 atom of Calcium, atomic weight (again according to Wikipedia)
40.08. 
This means we have 40.08*(3.668/147.02) g/l of calcium.  Multiply by 1000 for mg, and take that value in mg/l as equal to ppm.
 

Offline chloeukc

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Working out the concentration of a stock solution..
« Reply #2 on: 30/10/2007 09:26:11 »
That was literally what was written for me so that was all I had to play with! Thankyou for your help.
 

Offline techmind

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Working out the concentration of a stock solution..
« Reply #3 on: 30/10/2007 10:08:28 »
A stock solution of calcium was prepared by dissolving 3.668 g CaCl2.2H2O in water and diluting to 1000 cm3. What is the concentration of calcium in this stock solution in ppm?
I think you'll find that you need to take into account the water of crystallisation in your CaCl2
You've got less than 3.668g of CaCl2 because the weight given also includes the .2H2O.
I haven't run through the numbers, but this could easily make a factor of two or more difference in your answer.

Often with these kind of questions (in British exams at least) the numbers are contrived to make the answers "neat". If you follow the correct method, you might expect to find that the "3.668g" actually simplifies your working at the second or third step!


Edit: eric l has actually used the molar mass for the dihydrate straight from Wikipedia, so his working should actually be ok. For any kind of test however, you'd be expected to calculate the molar masses from scratch, and you'd be wise to explicitly account for the WoC.
« Last Edit: 30/10/2007 10:13:27 by techmind »
 

Offline eric l

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Working out the concentration of a stock solution..
« Reply #4 on: 30/10/2007 16:12:12 »
[Edit: eric l has actually used the molar mass for the dihydrate straight from Wikipedia, so his working should actually be ok. For any kind of test however, you'd be expected to calculate the molar masses from scratch, and you'd be wise to explicitly account for the WoC.
I took it for granted that with an online test, you had access to all sources on the internet, including WikipÍdia - just like we in our times had access to all catalogues of chemical compounds and even to the labels on the bottles with an "open book" test.
Of course, it does not harm to check the Wikipedia information with other sources (on the web or elsewhere).  Personally, I would go for the Material Safety Data Sheets - if available.
 

Offline Bored chemist

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Working out the concentration of a stock solution..
« Reply #5 on: 30/10/2007 20:06:45 »
CaCl2.2H2O has a molecular weight of (I don't know- look it up)- about 145
Of that only part is the calcium. Each mole of the salt contains one mole of Ca which is (I can't remember off hand, about 40g)

So 130 g of the salt is equivalent to about 40g of Ca
You took 3.668 g of the salt so you get about 40/145*3.668 g of calcium
Thats quite a lot of miligrams ( I will let you multiply by 1000 to convert g to mg.
A solution containing 1 mg in 1 litre of water is usually refered to as 1ppm (I know it's not precise because the density of the solution isn't exactly 1).
Your solution will contain  however many mg of Ca you worked out in 1 litre so it's that many ppm.

 

Offline chloeukc

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Working out the concentration of a stock solution..
« Reply #6 on: 30/10/2007 21:38:46 »
I'm clearly not cut out for analytical chemistry, as I cant work out how to do this one either:

A urine sample of a man suspected of suffering from chromium poisoning was analysed by atomic emission spectroscopy.  Chromium standards were prepared at concentrations of 5, 10, 15 and 20 ppm. To 100 cm3 of each standard and to 100 cm3 of the urine sample was added 0.1 cm3 of a 500 ppm yttrium solution. Emission from chromium and yttrium in each solution was then measured with the following results.

Solution          Chromium Emission (arbitrary units)     Yttrium Emission (arbitrary units)
5.0 ppm standard      13.0                                                      6.0
10.0 ppm standard      28.6                                                      6.6
15.0 ppm standard      31.2                                                      4.8
20.0 ppm standard      46.8                                                      5.4
Urine                      31.1                                                      5.2

Determine the concentration of chromium in the urine sample in ppm to one decimal place.
« Last Edit: 30/10/2007 21:40:30 by chloeukc »
 

Offline Bored chemist

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Working out the concentration of a stock solution..
« Reply #7 on: 31/10/2007 20:47:53 »
OK, a couple of questions; if it didn't mention yttrium would you know how to do it, and do you know what the yttrium is for?
 

Offline chloeukc

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Working out the concentration of a stock solution..
« Reply #8 on: 01/11/2007 10:08:47 »
No to both!!!
 

Offline Bored chemist

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Working out the concentration of a stock solution..
« Reply #9 on: 01/11/2007 19:55:30 »

 

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