[puccalie]

it is already clear that the definition of molarity is equivalent with the mole of solute (the component of solution that is dissolved to solvent, which usually in the form of solid or by definition is the smaller amount compare to the solvent) per litre of solution.

my question is :how if the solute is in the form of polymer-which is sticky (dense liquid) with the concentration is already known?

is it possible to prepare the solution by adding 1 litre  solution into an appropiate volume of polymer. or should i prepare by thinking as diluting the polymer into solution, following M1V1=M2V2  formula.

as a note,the polymer i treat is Ti butoxide tetramer, and the solution is butanol.

thankyou for helping

[barcrphd]

molarity of a solution is the "number of moles of the solutes dissolved in one litre of the solvent".

if i want to prepare 1Molar solution of sodium hydroxide, i will dissolve 40 gms of sodium hydroxide in 1L of water.

in a similar way, if you want to prepare 1M solution of your polymer and x is the molecular weight of your polymer, then you have dissolve x gms of polymer in 1L of appropriate solvent. in this case, butanol.

ofcourse, in the case of polymer you will be most probably using average molecular weight, since you cannot have a polymer with all the molecules wth the same molecular weight.

you need not confuse with the volume of the solute in case the solute is a liquid. you can go by the weight.

[puccalie]

thank you very much. This answer helps so much.

At first, I was just wondering because the total volume will be completely different when I just prepare 1mole of solute in 1 L solvent with the one when I tried to dilute following M1V1=M2V2 formula. The first one will have larger total volume compared with the last one.

Again, thankyou.