[camptzak]

If you are given the temperature at which an alcohol/water mixture boils, lets say 85°C, can you calculate the mol fractions strictly based on that temperature?

Ethanol boils at 78°C and water boils of course at 100°C

Raoult's Law: Ptotal = Pethanol + Pwater

Ptotal= 760torr

Pethanol= (mol fraction)(P°)        Pwater=(mol fraction)(P°)

assuming you had a table for the pressures of water an ethanol at different temperatures, could you calculate the mol fractions using (X) and (1-X) or something like that?

 

[Borek]

You can use tables, you can read the value from the boiling-point diagram for the water-ethanol mixture.

[curiouscat]

If you are given the temperature at which an alcohol/water mixture boils, lets say 85°C, can you calculate the mol fractions strictly based on that temperature?

Yes.

assuming you had a table for the pressures of water an ethanol at different temperatures, could you calculate the mol fractions using (X) and (1-X) or something like that?

Yes.

[camptzak]

Ok thanks guys, I worked a couple of those out. I know you can use the tables, but I am doing this more to see if I can do it algebraically, just for fun.

Heres a good one,

If you have known mol fractions, can you calculate the bp of the liquid?

X_toluene=0.3
X_Cyclohexane=0.7

Raoult's Law

P_total= P_cyclohexane+P_toluene

P_total= (0.7)(P°)+(0.3)(P°)

it looks like the temperature pressure curves are expressible as quadratic equations.
lets say for the questions sake
toluene: Y₁= X²+2X
Cycolhexane: Y₂=2X²+3X

so if Y₁+Y₂=760 then

(0.7)(2X²+3X)+(0.3)(X²+2X)=760

I just made up these equations for the pressure temperature relationship, so I have no way of verifying if this is the correct way to attack this problem.

[curiouscat]

Approach looks ok to me.

You do realize that mixtures have a Bubble / Dew point and not a single BP?

[camptzak]

Approach looks ok to me.

You do realize that mixtures have a Bubble / Dew point and not a single BP?

I did not know what, I don't even know what a dew point is... I am looking at wikipedia now ha. thanks