C1V1 = C2V2
Moderators: honeev, Leonid, amiradm, BioTeam
C1V1 = C2V2
Hi, I am lost with this. I want to prepare a 10ml of a 20?g/mL working solution, by diluting 0.1% w/v stock solution. Using C1V1 = C2V2
Does any know how to do this? Or have a website where the method is shown? thanks
Does any know how to do this? Or have a website where the method is shown? thanks
 Katy_Bobbles
 Garter
 Posts: 42
 Joined: Tue May 22, 2007 4:23 pm
 Location: Glasgow
Re: C1V1 = C2V2
0.1% is equivalent to 0.1g/100ml which is equivalent to 0.001g/ml. To make it an easier number to work with I would convert the units to mg/ml so C1 = 1mg/ml
Well C2 is your final concentration value so C2 = 20 (g/ml). But you also have to convert this to mg/ml = 20000
V2 is your final volume so that would be 10 (ml)
You want to find out the V1 so:
V1 = C2 X V2/C1
= (1) x 10/20000
= 0.0005 ml = 0.5μL
Thats the way I'd have done it. Is that the kind of value your looking for?? It's tiny in a 10ml final volume. Or is the question mark here:
supposed to be μg/ml?? Or am I just being stupid n have wrote a lot of rubbish?!?
Well C2 is your final concentration value so C2 = 20 (g/ml). But you also have to convert this to mg/ml = 20000
V2 is your final volume so that would be 10 (ml)
You want to find out the V1 so:
V1 = C2 X V2/C1
= (1) x 10/20000
= 0.0005 ml = 0.5μL
Thats the way I'd have done it. Is that the kind of value your looking for?? It's tiny in a 10ml final volume. Or is the question mark here:
dan167 wrote:Hi, I am lost with this. I want to prepare a 10ml of a 20?g/mL working solution, by diluting 0.1% w/v stock solution. Using C1V1 = C2V2
supposed to be μg/ml?? Or am I just being stupid n have wrote a lot of rubbish?!?
I assume you mean to prepare 10 ml of a 20 microgram/ml solution from a 1 mg/ml (0.1% w/v) stock. You can’t make a 20 g/ml solution by diluting a stock solution of only 0.001 g/ml, so there’s something wrong with Katy’s reasoning, though the gist of the explanation is correct.
The idea of using V1 x C1 = V2 x C2 is simple algebra (well, that and the principle of mass balance), but the trick is to make sure all your units are consistent, as Katy was suggensting, otherwise you may end up numerically off by 3 or more orders of magnitude. Let V1 = the unknown volume of stock solution needed, C1 = 0.1%(w/v) = 1 mg/ml, as Katy said, but I’ll take it 3 orders of magnitude lower and restate it as 1000 microgram/ml. Then V2 is the final volume of 10 ml of a C2 = 20 microgram/ml solution. Solve the expression for the unknown volume, V1 and you should have:
V1 = (V2 x C2)/C1.
Plug in the numbers and you get:
V1 = (10 ml x 20 microgram/ml)/ 1000 microgram/ml = 0.2 ml
So dilute 0.2 ml of 0.1% (w/v) stock to a final volume of 10 ml.
The idea of using V1 x C1 = V2 x C2 is simple algebra (well, that and the principle of mass balance), but the trick is to make sure all your units are consistent, as Katy was suggensting, otherwise you may end up numerically off by 3 or more orders of magnitude. Let V1 = the unknown volume of stock solution needed, C1 = 0.1%(w/v) = 1 mg/ml, as Katy said, but I’ll take it 3 orders of magnitude lower and restate it as 1000 microgram/ml. Then V2 is the final volume of 10 ml of a C2 = 20 microgram/ml solution. Solve the expression for the unknown volume, V1 and you should have:
V1 = (V2 x C2)/C1.
Plug in the numbers and you get:
V1 = (10 ml x 20 microgram/ml)/ 1000 microgram/ml = 0.2 ml
So dilute 0.2 ml of 0.1% (w/v) stock to a final volume of 10 ml.
C1V1=C2V2
Can someone please help me with this problem? If you could help in layman terms, I think I will be able to get the rest of my assignment. It's been a long time since I've done any type of chemistry. Thanks  here goes:
Given a stock solution of 5.0% sodium chloride (NaCl), how would you prepare 20 ml of the following solution?
2.0% sodiwm chloride solution.
Can someone please help me with this problem? If you could help in layman terms, I think I will be able to get the rest of my assignment. It's been a long time since I've done any type of chemistry. Thanks  here goes:
Given a stock solution of 5.0% sodium chloride (NaCl), how would you prepare 20 ml of the following solution?
2.0% sodiwm chloride solution.

 Garter
 Posts: 1
 Joined: Sat Apr 10, 2010 6:38 am
Re: C1V1 = C2V2
for the preparation of 2% NaCl (2g/100)from the 5% NaCl (5 g/100ml)
please apply the formul C1V1=C2V2
C1= 5 g
V1= requred to know
C2= 2 g
V2 = 20
put the values in the following formula
V1=C2 X V2 /C1
V1=2 X 20 / 5
V1= 8 ml
hence tkae the 8 ml solution from the stock solution of 5% NaCl and make up to 20 ml with distilled water.
by this way you can make 2% NaCl solution from 5% NaCl stock solution.
please apply the formul C1V1=C2V2
C1= 5 g
V1= requred to know
C2= 2 g
V2 = 20
put the values in the following formula
V1=C2 X V2 /C1
V1=2 X 20 / 5
V1= 8 ml
hence tkae the 8 ml solution from the stock solution of 5% NaCl and make up to 20 ml with distilled water.
by this way you can make 2% NaCl solution from 5% NaCl stock solution.
Who is online
Users browsing this forum: No registered users and 5 guests