Chemistry question. (I'm looking at you DTVT)

[73109]

So, I did a lab a few days ago and I was able to calculate the molar mass of antifreeze using the freezing point depression. It was a longish process and the very last question asked me to derive an equation that would find the molar mass of anything dissolved in water if given the freezing point depression and the mass of the solute. This is what I came up with and I think it is good, but I'm not sure if I could simplify it further:

Mass/      (.001∆Tmass)
            ___________
                  1.86

Anybody know if that is simplified to its fullest extent?

       
 

[Ravenheart]

aSoul + bTechnicality = Skill Level

[mizzl]

aSoul + bTechnicality = Skill Level

I smell a meme

[mizzl]

Mass/

*\ssaM

[ZBomber]

Did you try using science?

[rumborak]

Did you try using science?

Be careful, it can blind you.

rumborak

[millahh]

The great thing about being a chemist is that I don't have to remember how to do this kind of stuff anymore.

[mizzl]

A chemist once became a doctor. His patients soon noticed he would either Helium or Barium

[Dr. DTVT]

So, I did a lab a few days ago and I was able to calculate the molar mass of antifreeze using the freezing point depression. It was a longish process and the very last question asked me to derive an equation that would find the molar mass of anything dissolved in water if given the freezing point depression and the mass of the solute. This is what I came up with and I think it is good, but I'm not sure if I could simplify it further:

Mass/      (.001∆Tmass)
            ___________
                  1.86

Anybody know if that is simplified to its fullest extent?

       
 

Ok...remember how on facebook chat a few months ago I told you that melting point depression (meltinp point = freezing point) was not dependent on the material itself, just the molar amount of the material?  Well, this is what you will be showing.

dT = K*m*i  where:

dT is the depression observed
K is the cryoscopic constant for the solvent
m is the molality of the material
i is the mizzl (van't Hoff factor really), which is an integer which represents the number of molecules per mole

so, using your original data, you can solve for K, which will remain constant as long as you stay in water.

You can now rearrange to:

dT/(K*i) = m

I have to go to the airport and then take someone to dinner.  If you need further help, check your PMs.

[lonestar]

Mass/

*\ssaM

[glaurung]

A chemist once became a doctor. His patients soon noticed he would either Helium or Barium

Worst joke of the year.

[Volk9]

Think with portals...?

[slycordinator]

i is the mizzl (van't Hoff factor really), which is an integer which represents the number of molecules per mole

I'm definitely no chemistry expert (don't know what a van't Hoff factor is) but your description of it sounds more like Avogadro's number.

[Dr. DTVT]

i is the mizzl (van't Hoff factor really), which is an integer which represents the number of molecules per mole

I'm definitely no chemistry expert (don't know what a van't Hoff factor is) but your description of it sounds more like Avogadro's number.

What it describes is a deionization process.

i = 1 for something like sugar, which dissolves but does not break into ions
i = 2 would be for something like table salt, which breaks down into Na+ and Cl- ions

[Aramatheis]

i is the mizzl (van't Hoff factor really), which is an integer which represents the number of molecules per mole

I'm definitely no chemistry expert (don't know what a van't Hoff factor is) but your description of it sounds more like Avogadro's number.

Avogadro's is just the number of molecules (read; particles/atoms) per mole of substance.

van't Hoff factor is precisely this:

What it describes is a deionization process.

i = 1 for something like sugar, which dissolves but does not break into ions
i = 2 would be for something like table salt, which breaks down into Na+ and Cl- ions

[Tick]

Did you try using science?

Be careful, it can blind you.

rumborak

Only if you use your right hand.