Calculate Normal Boiling Point of Ethyl Alcohol

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SUMMARY

The normal boiling point of ethyl alcohol was calculated based on a solution of 26.0g of glucose dissolved in 285g of ethyl alcohol, resulting in a boiling point of 79.1°C. Using the ebullioscopic constant (Kb) of 1.22°C kg/mol for ethyl alcohol, the molality was determined to be 0.491 mol/kg. The change in boiling point (ΔTb) was calculated as 0.599°C, leading to a final normal boiling point of 79°C after correcting for the boiling point elevation. The calculations indicate a minor rounding error in the initial attempt.

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Homework Statement


What is the normal boiling point in Celsius of ethyl alcohol if a solution prepared by dissolving 26.0g of glucose C6H12O6 in 285g of ethyl alcohol has a boiling point of 79.1 Celsius? Kb for ethyl alcohol is 1.22( Celsius x kg)/mol .


Homework Equations



26g of glucose Mole conversion.
285 gram ----> kg
Molality, m = mole/kg

Equation = (Delta) Tb= Kb x m
Kb is given. m is what I found.

The Attempt at a Solution


Mole of 26g glucose is 26/180g (molar mass)= .14 moles
285g equals .285k kg
.14 mol/.285 kg = .491 m
---------------------------------------
Kb x m -------------> 1.22 x .491 = .599 Celsius
Delta T: 79.1 celsius - .599 celsius = 79 Celsius

My calculation has a rounding error? I'm close with the answer, idk what I'm doing wrong.
 
Physics news on Phys.org
Try subtracting 0.599 from 79.1 again.
 

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