Question

A 25.0g sample of glucose, C6H12O6, dissolves in 525mL of distilled water.

a. what is the boiling point elevation of the solution over the pure solvent?
b. At what temperature will the solution boil at sea level?

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I'm not a fan of these types of questions. This is the work I did for A:

boiling point elevation is
deltaT = i*Kb*m
i=7 kb=.512
moles of solute is 25g/180gpermole = .139 moles of glucose

i believe it's assumed that mL is grams, so 525 mL = 525g

and because (i believe) you need molality of solvent, this is what i have:

mols solvent/kg solvent
= 29mol water/.525kg water = 55.56

and finally, in the actual formula:

deltaT = 7(.512)(55.56)
= 199.11

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For some reason this answer doesn't seem right. What am I doing wrong here? Also, how do you do part B?

Thank you so much!

Answer 1

I don't know how you got i=7 lol
Way off - do you even know what i is?

Answer 2

you have everything right except you molality is moles of solute over kg of solvent so it would be .139/.525 for molality.

Answer 3

I is 1. This is a covalent molecule