# 10 millimolar (mM) to 250 micromolar - (Aug/28/2012 )

ascacioc on Wed Aug 29 18:49:09 2012 said:

useful formulae:

concentration (in M)= mass (in grams)/MW (in grams/mol)/volume (in L); since you know your concentration, volume and MW --> mass of substance in grams = volume (in L) * MW (in g/mol) * concentration (in M). This is your stock solution (10 mM), or the most concentrated from which you can take a volume V1 to add to V2 of media to get 200 uM final conc

c1*V1=c2*(V2+V1); where c1 is 10 mM = 10000 uM; c2 is 200 uM and V2 is the final volume on top of which you add your drug (~12 mL for a small flask of cells)

solve for V1 = c2 * V2 /(c1-c2). since you plug in V2 in mL, you will get V1 also in mL. (for V2 = 12 mL you get ~0.245 mL = 245 uL). For 20 uM you need to add 10 times less --> 24.5 uL or you dilute your stock solution 10 times (1 part stock of 10 mM and 9 parts media) (down to 1 mM) and still add 245 uL.

With the two formulae I gave you, you can also answer the second question.

Link to similar discussion with drug soluble in DMSO:

http://www.protocol-...-added-to-dmem/

http://www.protocol-...__hl__micronagu

Andreea

I am getting confused with the formula used and where did i get v2- 12 ml. I dont think this is making any sense to me.

cell_bee on Thu Aug 30 20:49:23 2012 said:

ascacioc on Wed Aug 29 18:49:09 2012 said:

useful formulae:

concentration (in M)= mass (in grams)/MW (in grams/mol)/volume (in L); since you know your concentration, volume and MW --> mass of substance in grams = volume (in L) * MW (in g/mol) * concentration (in M). This is your stock solution (10 mM), or the most concentrated from which you can take a volume V1 to add to V2 of media to get 200 uM final conc

c1*V1=c2*(V2+V1); where c1 is 10 mM = 10000 uM; c2 is 200 uM and V2 is the final volume on top of which you add your drug (~12 mL for a small flask of cells)

solve for V1 = c2 * V2 /(c1-c2). since you plug in V2 in mL, you will get V1 also in mL. (for V2 = 12 mL you get ~0.245 mL = 245 uL). For 20 uM you need to add 10 times less --> 24.5 uL or you dilute your stock solution 10 times (1 part stock of 10 mM and 9 parts media) (down to 1 mM) and still add 245 uL.

With the two formulae I gave you, you can also answer the second question.

Link to similar discussion with drug soluble in DMSO:

http://www.protocol-...-added-to-dmem/

http://www.protocol-...__hl__micronagu

Andreea

I am getting confused with the formula used and where did i get v2- 12 ml. I dont think this is making any sense to me.

I forgot all about your question.

Just forgot that formula ... there is no point in it if you dont understand the math/logic behind it.

I'll try to answer your question in the following 2days.

Anyway: I want you to write down 2 things => first you write down everything you know (write down what you know with the correct units and what it means + the meaning of those things, what are moles ? What is MW ? What do you do to go from not diluted to 10^-3) and secondly: write down what you need (need to calculate).

I took 12 mL as an example of final volume (totally random) As Pito says: state everything you know about your system: final volume would be something important to mention

**plate count**

**dilution factor**

**volume added to plate**

**overall dilution factor**

**sample concentration**

250

10^{6}

0.1

10^{7}

2.50 x 10^{9}

Reference-- http://www.mansfield.ohio-state.edu/~sabedon/biol4039.htm

C1V1 = C2V2

2.50 x 10^{9 }* X = 10^{6 }* 10ml

X = 10^7/ 2.5* 10^9

= 0.4 * 10^-2

= 0.004ml

= 4 ul

Dilution factor is written as 10^6 with positive and concentration factor as 10^-6

prabhubct on Fri Aug 31 06:46:21 2012 said:

**plate count**

**dilution factor**

**volume added to plate**

**overall dilution factor**

**sample concentration**

250

10

^{6}

0.1

10

^{7}

2.50 x 10

^{9}

Reference-- http://www.mansfield...on/biol4039.htm

C1V1 = C2V2

2.50 x 10

^{9 }* X = 10

^{6 }* 10ml

X = 10^7/ 2.5* 10^9

= 0.4 * 10^-2

= 0.004ml

= 4 ul

Dilution factor is written as 10^6 with positive and concentration factor as 10^-6

know what you mean, but you are not reading what I wrote down.

You state:

*"and 10^7 must be more*concentrated original stock than 10^9 because its serially diluted"

and at the same time:

*This means 10^6 diltion stock is of*2.5*10^9 CFU/mL.

10^5 diltion stock is of

2.5*10^8 CFU/mL.

*10^4 diltion stock is of*

*2.5*10^7 CFU/mL.*

10^4 diltion stock is of

2.5*10^6 CFU/mL.

10^3 diltion stock is of

2.5*10^5 CFU/mL. This does not add up.

If the dilution factor is written with a positive power then both your sentences are about dilution factors? And thus you are making a mistake in one of them.

==> the higher the dilution factor, the lower the concentration!

Or?

It was mistake.

I think

*This means*

*10^9 dilution stock is of **2.5*10^6 CFU/mL.*

*10^8 dilution stock is of **2.5*10^7 CFU/mL.*

*10^7 diltion stock is of **2.5*10^8 CFU/mL.*

*10^5 dilution stock is of **2.5*10^10 CFU/mL.*

*10^4 dilution stock is of **2.5*10^11 CFU/mL.*

*10^3 dilution stock is of **2.5*10^12 CFU/mL.*

*10^8 dilution stock is of **2.5*10^7 CFU/mL.*

Add 40ul of *10^8 dilution stock is of **2.5*10^7 CFU/mL. to 9960ul to make upto 10ml.*

For cross checking check below,

*2.5*10^7 CFU/ml ------ 1000 ul*

*so Y ------ 40 ul*

*Y = 10^9/ 10^3 *

*= 10^6CFU/ml*

prabhubct on Fri Aug 31 10:25:38 2012 said:

It was mistake.

I think

*This means*

*10^9 dilution stock is of*

*2.5*10^6 CFU/mL.*

*10^8 dilution stock is of*

*2.5*10^7 CFU/mL.*

*10^7 diltion stock is of*

*2.5*10^8 CFU/mL.*

10^6 dilution stock is of 2.5*10^9 CFU/mL.

*10^5 dilution stock is of*

*2.5*10^10 CFU/mL.*

*10^4 dilution stock is of*

*2.5*10^11 CFU/mL.*

*10^3 dilution stock is of*

*2.5*10^12 CFU/mL.*

*10^8 dilution stock is of*

*2.5*10^7 CFU/mL.*Add 40ul of

*10^8 dilution stock is of*

*2.5*10^7 CFU/mL. to 9960ul to make upto 10ml.*For cross checking check below,

*2.5*10^7 CFU/ml ------ 1000 ul*

*so Y ------ 40 ul*

*Y = 10^9/ 10^3*

*= 10^6CFU/ml*Ok, I understand this.

But you see now how difficult it can be if you dont define what you mean from the start and if you dont use the powers correctly.

pito on Fri Aug 31 12:42:48 2012 said:

prabhubct on Fri Aug 31 10:25:38 2012 said:

It was mistake.

I think

*This means*

*10^9 dilution stock is of*

*2.5*10^6 CFU/mL.*

*10^8 dilution stock is of*

*2.5*10^7 CFU/mL.*

*10^7 diltion stock is of*

*2.5*10^8 CFU/mL.*

10^6 dilution stock is of 2.5*10^9 CFU/mL.

*10^5 dilution stock is of*

*2.5*10^10 CFU/mL.*

*10^4 dilution stock is of*

*2.5*10^11 CFU/mL.*

*10^3 dilution stock is of*

*2.5*10^12 CFU/mL.*

*10^8 dilution stock is of*

*2.5*10^7 CFU/mL.*Add 40ul of

*10^8 dilution stock is of*

*2.5*10^7 CFU/mL. to 9960ul to make upto 10ml.*For cross checking check below,

*2.5*10^7 CFU/ml ------ 1000 ul*

*so Y ------ 40 ul*

*Y = 10^9/ 10^3*

*= 10^6CFU/ml*Ok, I understand this.

But you see now how difficult it can be if you dont define what you mean from the start and if you dont use the powers correctly.

Yeah Pito, Its quiet simple dilution problem. just I missed power (-Ve, +ve confusion) and some wrong math in hurry. your critical remarks do helped a lot.

* *

Do I need to take 4

chandch on Fri Aug 31 17:00:46 2012 said:

Do I need to take 4

Yes.

You are right.

However 1 more question: would you really do it like this?

Would you add 4µl into 9,996 ml?