Answer :
Final answer:
The boiling point of the solution will be 0.57 ℃ higher than the boiling point of pure water. Therefore, the solution will begin to boil at the boiling point of pure water (100 ℃) plus the boiling point elevation (0.57 ℃), giving a boiling point of approximately 100.57 ℃.
Explanation:
Calculating boil point(temperature):
The boiling point of a solution is determined by the concentration of solute particles in the solvent. In this case, we have a solution consisting of glucose (C6H12O6) in water (H2O). The boiling point elevation can be calculated using the equation ΔTb = iKbm, where ΔTb is the boiling point elevation, i is the can't Hoff factor, Kb is the molal boiling point elevation constant, and m is the molality of the solution.
For glucose in water, the van't Hoff factor (i) is 1 and the molal boiling point elevation constant (Kb) for water is approximately 0.512 ℃·kg/mol. Since we know the number of moles of glucose (0.73 moles) and the mass of water (650 mL = 0.65 kg), we can calculate the molality (m) of the solution as follows:
molality (m) = moles of solute / mass of solvent = 0.73 moles / 0.65 kg = 1.12 mol/kg
Using the equation ΔTb = iKbm, we can solve for the boiling point elevation:
ΔTb = (1)(0.512 ℃·kg/mol)(1.12 mol/kg) = 0.57 ℃
The boiling point of the solution will be 0.57 ℃ higher than the boiling point of pure water. Therefore, the solution will begin to boil at the boiling point of pure water (100 ℃) plus the boiling point elevation (0.57 ℃), giving a boiling point of approximately 100.57 ℃.
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