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Torshi
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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
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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.