Millimoles conversion to mL or grams?

[Alex]

Hello,

I have a question regarding milli moles conversion.

How much approx. would a 70 milli molar (mM) solution of DMSO be in mL and what volume of pure 99.9% DMSO be required to create a 70 mM solution?

What formula is required for conversion from milli moles (mM) to grams/mL?

Thanks.



Edited:

Unable to change the thread heading, from Micro moles to Milli moles conversion to mL or grams?

Mod Edit: I am able to change it

[JohnS]

You need the molar mass of the substance. The molar mass is numerical equal in grams to the atomic weight of the molecule. You can often look it up in Wikipedia, or from the chemical formula, you can add up all the atomic weights of the atoms in a molecule. Looking it up, molar mass is 78.13 g/mol. If you want to work in volumes, you also need density, 1.1 g/mL.

M (Molar)is a shorthand for molles per liter (mol/L). So if you wanted 1 L of 70 mM solution, you need 70 mmol per liter of solution.

70 mmol x 78.13 g/mol = 5.469 g. Using the density, that is 4.972 mL of DMSO per liter of solution you wish to make.

[Alex]

Thanks for the expert detailed explanation!

Best regards!

[Alex]

Since it's 4.972 mL of DMSO in 1 L (of saline/water?), what would be the amount in 1 mL, 0.004972 mL of 70 milli molar (mM) DMSO...correct?

If yes, than 100 mL of water would contain 0.4972 mM of DMSO?

What would be the amount in mM in a 99.9% DMSO?

Thanks again.

[JohnS]

Since it's 4.972 mL of DMSO in 1 L (of saline/water?), what would be the amount in 1 mL, 0.004972 mL of 70 milli molar (mM) DMSO...correct?

If yes, than 100 mL of water would contain 0.4972 mM of DMSO?

What would be the amount in mM in a 99.9% DMSO?

Thanks again.

100 mL would require 0.4972 mL not millimolar.

If the DMSO is only 99.9% pure, divide any of the above numbers by 0.999, making them roughly 1.001 times the quantities above.

[Alex]

100 mL would require 0.4972 mL not millimolar.

If the DMSO is only 99.9% pure, divide any of the above numbers by 0.999, making them roughly 1.001 times the quantities above.

100 mL would contain 0.4972 mL and NOT mM.
1,000 mL would contain approx. 491.2 mL of 99.9%, 70 mM.
Hypothetically let me assume it has absorbed some moisture being hygroscopic, and now is 99% or 98%.
Divide 491.2 by 1/2 = 490.7/245.6 mL of solute. Seems a bit high. It would make almost a near 50% net solution.

[Alex]

Thanks anyways, appreciate your expert prompt reply. My only simple question, which still is bothering (since am not an organic chemist and bad at math's) is how much 99.9% of the DMSO solute do I need to add to water or saline, to make a net solution of 70 milli moles (mM), what formula do I use to make a net solution from 30-140 mM.

Impurities in the solute can be overlooked, so can the volume in which is being added. To know the approx. net volume of the solute. I have a 99.9999% purity solute too. Just need a simple formula for the approx. solute for a given mM.

Thanks again.

[Alex]

Lastly I have used 70 mL of the solute in 900 mL of saline that's approx. 7.77& by volume. What would be the content of solute in mM in it?

[LiamHermann]

Thanks for answering, you made my day. Now I won't get punished by my teacher.