Question

determine the boiling points of 1M solutions of sugar,glucose,urea,sodium chloride,barium chloride,aluminium chloride

Answer 1

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

Answer 2

hi bubby

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