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Yakvenalex [24]
3 years ago
6

What is the boiling point of a 1.5 m aqueous solution of fructose? the boiling point elevation constant of water is 0.515°c/m.

assume solute is a nonelectrolyte?
Chemistry
1 answer:
Ronch [10]3 years ago
4 0
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.<span>
ΔTb -  the boiling point elevation.
Kb - the ebullioscopic constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
 Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
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We have the mass of each reactant, so let's convert the mass to moles:

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