What is the boiling point elevation of a solution that contains 20.0 grams of glucose (C6H12O6) in 1000.0 grams of water?
Correct: A
To calculate the boiling point elevation, we need to calculate the molality of the solution. First, we calculate the number of moles of glucose: moles = mass / molar mass = 20.0 g / 180.16 g/mol = 0.111 mol. Then, we calculate the molality: molality = moles / mass of solvent (in kg) = 0.111 mol / 1.00 kg = 0.111 m. The boiling point elevation constant (Kb) for water is 0.512°C/m. Therefore, the boiling point elevation is ΔTb = Kb * m = 0.512°C/m * 0.111 m = 0.057°C, which is closest to 0.10°C among the given choices.