Deterimination of the boiling point of the 1 molamass of the glucose,sodium chloride,berium chloride
Answers
We assume the solvent is water. Normal boiling point at 1 atm is 100°C.
We assume that the pressure is constant.
Raoults law: ΔT_b = K_b * i * b_solute
K_b = R (T_b)² M / ΔH_v = Ebullioscopic constant (for a solvent)
R: Universal gas constant
T_b = boiling point of pure solvent
M = Molar mass
ΔH_v = Latent Heat of vaporization
K_b = 0.512 °C /molal for water
i = number of solute particles in solution (dissociated ions or single compound) per molecule
b_solute = molality of solute = number of moles of solute/kg of solvent
= 1 given
ANSWER:
For Sugar, Glucose, Urea : i = 1, ΔT_b = 0.512 °C T_b = 100.512°C
For Na Cl : i = 2 (or 1.9) : ΔT_b = 1.024°C, T_b = 101.512 °C
For Ba Cl2 : i = 3 : T_b = 101.536 °C
for Al Cl3 : i = 4 T_b = 102.048°C