Boiling Point Determination

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
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Do the experiment!!!
Paul

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ravi wrote:

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 ---------
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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

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a) or unless the vapor pressure is increased by putting the solution in a pressure vessel.

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