My Math Forum Preparing a concentration before carrying out serial dilutions

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 April 11th, 2017, 09:03 PM #1 Member   Joined: May 2015 From: Australia Posts: 50 Thanks: 6 Preparing a concentration before carrying out serial dilutions Hi, the question is in the link below: https://www.dropbox.com/s/zj2soz1eqs...ketch.png?dl=0 I don't understand the part where I marked yellow. I asked my teacher and he wrote his working out: Dilution factor = C(initial) / C(final) 2=4/C(final) therefore, C (final)=2 2mL of 4 unit/mL from stock C1V1=C2V2 10(V1)=(4)(2) V1=0.8mL stock therefore, 1.2mL diluent In my teacher's working out, I don't understand how C(final) = 2 becomes 2mL from stock (the parts where I marked in bold) Any help would be appreciated! Last edited by skipjack; April 12th, 2017 at 03:16 AM.
 April 12th, 2017, 03:13 AM #2 Global Moderator   Joined: Dec 2006 Posts: 17,728 Thanks: 1359 Looking ahead, he's worked out that only 2mL of 4units/mL solution will be needed. To change the concentration by a factor of 4/10, the volume of diluent must be multiplied by 10/4. That means that for every 4 parts of stock solution you use, you must add 6 parts of diluent, and that will produce a volume of 10 parts of solution with a concentration of 4units/mL. By choosing 10 parts to mean a volume of 2mL, you would create 2mL of 4units/mL solution. That means the 4 parts of the stock solution you use is 0.8mL of that solution, and adding 6 parts of diluent is adding 1.2mL of diluent. Thanks from pianist
 April 12th, 2017, 04:25 PM #3 Member   Joined: May 2015 From: Australia Posts: 50 Thanks: 6 Thanks, I understand the diluting part now. The calculations are easy once I know that 2mL is needed. Could you explain how he figured out to use 2mL? In my previous post, I talked about how I asked the teacher to expand on his solution, and he wrote this: Dilution factor = C(initial) / C(final) 2=4/C(final) therefore, C (final)=2 It doesn't make sense, his calculations say Concentration (final) = 2 How does a concentration become 2mL? The full question is in the link below https://www.dropbox.com/s/zj2soz1eqs...ketch.png?dl=0 thanks Last edited by skipjack; April 13th, 2017 at 06:03 AM.
 April 13th, 2017, 06:46 AM #4 Global Moderator   Joined: Dec 2006 Posts: 17,728 Thanks: 1359 The concentration didn't become 2mL. He foresaw how the serial dilutions later described would mean that he first needs to prepare 2mL of 4 units/mL concentration solution. However, there is a small mistake in the description of the serial dilutions. At the end, all the concentrations are correct, but the 2mL of solution in tube E needs to be reduced to 1mL by discarding 1mL of it as a final step. Thanks from pianist
 April 19th, 2017, 12:58 AM #5 Member   Joined: May 2015 From: Australia Posts: 50 Thanks: 6 Thanks for replying. If my teacher foresaw that he would need to prepare 2mL of 4units/mL, what was the purpose of doing the calculations below? Dilution factor = C(initial) / C(final) 2=4/C(final) therefore, C (final)=2 One more question, how did the teacher 'know' that he would need to prepare 2mL of 4 units/mL? I think the 2mL contains 1mL diluent and 1mL stock, however why didn't the teacher use 0.5mL or 3mL or 4mL? Sorry for asking so many questions about this; my test is on Monday and I can't contact my teacher because we're on study break. Last edited by skipjack; April 19th, 2017 at 02:30 AM.
 April 19th, 2017, 03:33 PM #6 Global Moderator   Joined: Dec 2006 Posts: 17,728 Thanks: 1359 The problem requires the preparation of 1mL each of the solution at various concentrations. The specified successive concentrations are reduced by a factor of 2, which suggests the 2mL procedure described, where you make 2mL of each concentration, but then use 1 mL of that concentration to make 2mL of the next concentration. It's possible, however, to use a similar procedure that utilizes slightly less than 2mL, but the amount of diluent used is then different for each concentration required. It's more efficient to do that, but more complicated to describe and requires accurate measurement of obscure fractions of 1mL. I assume that the "2=4/C(final)" calculation was supposed to be stating the obvious, but actually tends to confuse. He seems to be referring to just the initial part of the procedure he gives. What matters is that if you double the volume by adding an equal volume of diluent, the concentration is halved. Thanks from pianist

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