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Dilution factor 50 ul in 2mls


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6 replies to this topic

#1 agorganic

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Posted 25 September 2012 - 03:44 AM

Hi there,

I'm wondering how to work out the dilution factor for

20 microliters (ul) in 2000 microliters (ul)

50 ul in 2 milliliters (ml)

100 ul in 2 ml

and

150 ul in 2 ml

In case I don't understand ur calculation, please put the dilution factors as well. I'm getting fairly tired and have alot of work ahead that is sapping my little mental strength that remains.

#2 Inbox

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Posted 25 September 2012 - 04:05 AM

20 microliters (ul) in 2000 microliters (ul)

Dilution factor = Total volume / sample volume
= (20+2000)/20
= 101



But if you are adding to make up the final volume to 2000 then

Dilution factor = Total volume / sample volume
= (20+1980)/20
= 100

Rest you can calculate yourself.

Edited by prabhubct, 25 September 2012 - 04:06 AM.


#3 drwho

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Posted 27 September 2012 - 12:04 PM

Small addition to the above post, make sure everything is in the same units. For example you state the dilutions as 100ul in 2ml it is much easier to think of as 100ul in 2000ul

#4 Inbox

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Posted 27 September 2012 - 11:28 PM

100ul in 20000ul

#5 agorganic

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Posted 01 October 2012 - 03:06 PM

Can I write it as 100:2000, as this format is what I'm referring to specifically.

Thank you.

#6 DRT

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Posted 02 October 2012 - 10:55 AM

Can I write it as 100:2000, as this format is what I'm referring to specifically.


It will often be written this way but that depends on your academic lineage. Chemists will more typically use the ratio symbol to designate the proportion of each component (1:19 in your case) and the division symbol to designate the proportion of the final volume (ie 1/20). This makes it easier when there are multiple solutions to be added.
There would be no room for confusion if you used the ‘in’ notation that drwho and prabhubct have above.

#7 mdfenko

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Posted 03 October 2012 - 07:30 AM

Can I write it as 100:2000, as this format is what I'm referring to specifically.

no. convention is to use lowest denominator while maintaining whole numbers.
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