Answer
The freezing point o the solution = 273.714153 K
The boiling point of the solution= 373.1955328 K
Step-by-step explanation
Given:
Mass of glucose = 5.00 g
Volume of water = 72.8 g
What to find:
The freezing point and the boiling point of the solution.
Step-by-step solution:
Note: (Freezing point of water = 273K, Kf for water =1.87K kg/mol, atomic weight C = 12, H = 1, O = 16).
The freezing point of the solution:
The molecular weight of glucose C6H12O6 = 6(12) + 12(1) + 6(16) = 180 g/mol
The number of moles of glucose = (Mass of glucose/Molecular weight) = 5.00 g/180.0 g/mol = 0.0278 moles
Mass of water = 72.8 g = 0.0728 kg
So molality of glucose = (Moles o glcsose/Volume of solution) = 0.0278 mol/0.0728 kg = 0.3819 mol/kg
The depression in the freezing point, ∆T = Kf x molality = 1.87 K kg/mol x 0.3819 mol/kg = 0.714153 K
Since the freezing point of water = 273 K
Therefore, the freezing point of the solution= 273 K + 0.714153 K = 273.714153K
The boiling point of the solution:
∆T = i x m x Kb
∆T = change in temperature i.e boiling point elevation
i = van't Hoff factor = 1 for glucose since it does not ionize or dissociate. It is a single particle.
m = molality = moles solut/kg solvent = 0.0278 mol/0.0728 kg = 0.3819 mol/kg
Kb = boiling poin constant = 0.512 K kg/mol
∆T = 1 x 0.3819 mol/kg x 0.512 K kg/mol
∆T = 0.1955328 K
Since the freezing point of water = 373 K
Therefore, the boiling point of the solution = 373 K + 0.1955328 K = 373.1955328 K