Chemistry, asked by aishuaishaishu7875, 1 year ago

Deterimination of the boiling point of the 1 molamass of the glucose,sodium chloride,berium chloride

Answers

Answered by shaikmohaseeno52
3

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


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