Calculate Normal Boiling Point of Ethyl Alcohol

AI Thread Summary
The normal boiling point of ethyl alcohol is calculated using the boiling point elevation formula, where the boiling point of a solution with 26.0g of glucose in 285g of ethyl alcohol is 79.1°C. The molality of the solution is determined to be approximately 0.491 m. Using the provided ebullioscopic constant (Kb) of 1.22°C kg/mol, the elevation in boiling point is calculated to be 0.599°C. Subtracting this value from the boiling point of the solution yields a normal boiling point of 79°C for ethyl alcohol. The discussion highlights a potential rounding error in the calculations but confirms the methodology used.
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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.
 
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Try subtracting 0.599 from 79.1 again.
 
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