Chromatography Forum

Hi,

I have (10ml flask) inside this flask is a chemical ( x) .This chemical was dissolved ( partly ) in 10% methanol and 90% water .

my question is that i need to transfer the whole amount ( 10ml which contains 10% methanol and 90% water ) to 100 ml flask but i need the concentration of methanol is the same 10% and the also the same 90% water .please ignore the chemical concentration because even if i transfer ut all to 100 ml it will not dissolve all because is in an excessive amount

Am i right if i do two ways :

1- take the whole a mount of 10 ml put it in 100 ml flask ,then add 9ml (100% ) pure methanol from the bottle and compleed with water to the mark.

2- is it true if i prepare seperate 10% methanol 90%water in new volumetric flask ,then i add the amount of 10 ml ,,, does this will make 10% methanol in water ??

thanks for your help

Neither will work, because of something called volume (or mixing ) contraction. There is an Textbook on Physical Chemistry (Atkins?) that describes a fraudulent barkeeper trying to falsify wodka and failing because of this contraction.

Btw, is your % mass or volume or molarity?

If it was just for chromatography, I would use proposal 2 and make sure that it is always done in exactly the same way.

Alex

I'm a little confused by the question, so let me offer my take, based upon my understanding:

Why can't you make up a 10% methanol/90% water solution, then use that for making up your standards and solutions ???

I'm a little confused by the question, so let me offer my take, based upon my understanding:

Why can't you make up a 10% methanol/90% water solution, then use that for making up your standards and solutions ???

well , it is confusion i agree , but i will tell you why i would like to do this .Because i supposed to prepare my chemical in a large volume ( say 100 ml ) and i need this chemical to be in a condition of ( 10% methanol , 90% water ) , by mistak i prepared it in 10 ml .So i am thinking now to transfer the whole a mount of 10 ml to the 100 ml and try to make the condition is the same 10% methanol . The chemical is very expensive so i don't want to lose it .

thanks for u

Neither will work, because of something called volume (or mixing ) contraction. There is an Textbook on Physical Chemistry (Atkins?) that describes a fraudulent barkeeper trying to falsify wodka and failing because of this contraction.

Btw, is your % mass or volume or molarity?

If it was just for chromatography, I would use proposal 2 and make sure that it is always done in exactly the same way.

Alex

thanks for you and i would be very gratefull if you can give the title of the book with author because i really need to understand well the concentration in general and specific

i really appreciate your comment , thank a again

I'm not sure if I understand properly, but I don't think there is an impossible problem here.

You have partially dissolved X in 10mL 10% MeOH, and you need it fully dissolved in 100mL 10% MeOH. I'd follow Consumer Products Guy:

make up an excess of 10% MeOH in a spare container
pour your original 10mL into the new 100mL flask
use your stock of 10% MeOH to wash the old flask thoroughly into the new, transfering all undissolved solid, and leaving a little short of the 100mL mark.
Stir the new until dissolved
Make up the new flask with 10% MeOH to an accurate 100mL

How did you prepare your initial 10 mL solution?

If you dissolved your sample in 1 mL of methanol and made up to volume with water.
or
If you dissolved your sample in 9 mL of water and made up to volume with methanol.

then ther is a possiblity that when mixed your final volume may not be 10 mL or you may not have and exact 90:10 mix.

however if you follow previous instructions and perform the dilution to 100 mL with a 90:10 mix, the ratio in the 100 mL should have a negligable difference from a "true" 90:10 mix.