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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.