Jump to content

  • Log in with Facebook Log in with Twitter Log in with Windows Live Log In with Google      Sign In   
  • Create Account

Submit your paper to J Biol Methods today!
Photo
- - - - -

Concentration/molarity question (apologies)


  • Please log in to reply
5 replies to this topic

#1 Peter Kullar

Peter Kullar

    member

  • Members
  • Pip
  • 2 posts
0
Neutral

Posted 11 October 2013 - 06:54 AM

Hi there,

 

Apologies for posting a basic questions.....

 

I have 25mg powder in my tube, mol weight = 258.23.  It needs to be reconstituted in water at min 7mg/ml.

 

I want to have a six well plate with 0.5mM, 1mM, 2mM and 5mM solutions in the wells.

 

By my calculation - need to dissolve 3.6ml water

 

Total moles in tube = 25*10-3 / 258.23 = 0.00096 moles

 For 0.5mM well, need 0.5*10-3 * 2*10-3 = 1*10-6 moles (assuming that I put in 2 ml of media in the well)

 

Therefore I need 1*10-6/0.00096 = 0.001* 3.6 = 3.5 microlitre of the original solution in 2 ml of media in the well to make 0.5mM 

 

is this right?



#2 phage434

phage434

    Veteran

  • Global Moderators
  • PipPipPipPipPipPipPipPipPipPip
  • 2,506 posts
253
Excellent

Posted 11 October 2013 - 08:13 AM

This is a confusing way of solving this. I'd strongly suggest that you make a standard solution. Since you will be working with molar concentrations, you might make it at 1 M.

To make this solution, you need to add x liters of water to your compound, making a solution of concentration of [ 25 mg/(258 g/mole) ]/x liter = 1 mole/liter.

x liter = [ 25 mg/(258 g/mole) ] / 1 mole/liter = .000969 liter = 967 ul.

 

Now, you can calculate the amount of this solution you need in each well easily. For a well with 0.5 mM, you need to dilute the solution 2000x, so you if your well has a volume of 2 ml, you need 1 ul. For a well with 1 mM, you need to dilute by 1000x, so the volume you would add to 2 ml is 2 ul.



#3 chandra3316

chandra3316

    member

  • Active Members
  • Pip
  • 17 posts
2
Neutral

Posted 11 October 2013 - 09:46 AM

ada



#4 chandra3316

chandra3316

    member

  • Active Members
  • Pip
  • 17 posts
2
Neutral

Posted 11 October 2013 - 09:59 AM

Please try to use the molarity calculator for your calculations. Do not make calculation complicated. the link is,

http://www.graphpad....olarityform.cfm

 

prepare stock for 20millimolar(25mg in 4.84ml)

for six well plate(2ml) add,

5mM=500ul

1mM=100ul

0.5mM=50ul

of stock to final 2ml volume in the well. If you want to decrease above stock volumes make stock concentrted(100 milli molar or higher)

chandra3316@gmail.com



#5 phage434

phage434

    Veteran

  • Global Moderators
  • PipPipPipPipPipPipPipPipPipPip
  • 2,506 posts
253
Excellent

Posted 11 October 2013 - 11:01 AM

If the only way you can solve a problem like this is to use a web based calculator, you have a problem.

Moreover, you won't be able to notice when the calculator is telling you to do something ridiculous.

Also, your solution above violates one of the constraints: that the stock solution be above 7 mg/ml.



#6 Peter Kullar

Peter Kullar

    member

  • Members
  • Pip
  • 2 posts
0
Neutral

Posted 11 October 2013 - 10:32 PM

This is a confusing way of solving this. I'd strongly suggest that you make a standard solution. Since you will be working with molar concentrations, you might make it at 1 M.

To make this solution, you need to add x liters of water to your compound, making a solution of concentration of [ 25 mg/(258 g/mole) ]/x liter = 1 mole/liter.

x liter = [ 25 mg/(258 g/mole) ] / 1 mole/liter = .000969 liter = 967 ul.

 

Now, you can calculate the amount of this solution you need in each well easily. For a well with 0.5 mM, you need to dilute the solution 2000x, so you if your well has a volume of 2 ml, you need 1 ul. For a well with 1 mM, you need to dilute by 1000x, so the volume you would add to 2 ml is 2 ul.

Thanks a lot, yes, this is a much better way of working it out. thanks for your time.  best wishes, Peter






Home - About - Terms of Service - Privacy - Contact Us

©1999-2013 Protocol Online, All rights reserved.