Hi All,
I am looking forward the Boiling Point of mixture of
Water+Ethanol
Water+Acetone
Water+ChCl3
Water+Methanol
Water+IsoPropanol
composition of each mixture is 5g/5g.
Please help me.
Thanks,
Ravi
Hi Ravi,
I think this can be calculated as follows:
delta_T = k_e * m_a
where k_e is the ebullioscopic constant of ethanol (in your first
example) and m_a the molality of ethanol.
m_a = m(ethanol) / M(ethanol)
where m(ethanol) is the mass of ethanol in 1 kg water.
(this is 1000 g since your mixture is 1:1)
==> m_a = 1000 [g/1 kg] / 46.07 [g/mol]
k_e(ethanol) = 1.04 K*kg/mol (looked up)
==> delta_T = 1.04 [K*kg/mol] * 1000 [g/kg] / 46.07 [g/mol]
==> delta_T = 22.5734 [K*kg*g*mol/mol*kg*g] = 22.5734 kelvin.
So your first mixture will boil at 122.5734 grade celsius.
Just replace k_e in this example by the correct values of your other
substances ( try a search engine...) to find the corresponding temperatures.
regards,
Christian
---------
I am no chemist but its a pity all that science was wasted. (a) Is it
not apparent that (a) such a mixture could never boil at a higher
temperature than either of the solvents unless a compound was formed? (b)
If it does boil at that temperature then how has civilisation ever learned
to make spirits (whiskey, rum, vodka etc). I thought that fermented
mixtures of sugars and flavourings were heated to a temperature at which the
ethanol (plus a few of the flavours) boiled off leaving a constant boiling
mixture of water and a very small amount of ethanol, far less concentrated
than the original fermented liquor and hugely less that 50/50. Hiccup,
oops, whash was that?
Bob
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.