Calculate the freezing point and the boiling point of each of the following solutions using the observed van’t Hoff factors in Table 10-6.

a. 0.050 m MgCl₂

b. 0.050 m FeCl₃

Interpretation: The freezing point and the boiling point of the given solutions are should be calculated.

Concept introduction:

• Elevation of boiling point:

  The boiling point of the solution is increases when the solute is dissolved in the solvent is called Elevation of boiling point. it is one of the colligative Properties thus,

                                           ΔT = iKbmsolute......(1)           ΔT is boiling-point elevation          Kb is molal boiling-point elevation constant          m is  molality of the solute          i is the van't Hoff factor

• Depression in freezing point:

  The freezing point the solution is decreases when the solute is dissolved in the solvent is called Elevation of boiling point. it is one of the colligative Properties thus

                                        ΔT = iKfmsolute......(2)           ΔT is boiling-point elevation          Kf is molal freezing-point depression constant          m is  molality of the solute          i is the van't Hoff factor

Record the given data,

           Molality of MgCl2= 0.050 m 

           Molality of FeCl3= 0.050 m 

To calculate the boiling point of 0.050 m MgCl2 solution.

Molal boiling-point elevation constant of water is 0.51 °C/molal .

                          van't Hoff factor of MgCl2 is 2.7ΔT = 2.7 ×0.51 °C/molal× 0.050 molal       = 0.069°C      = 100°C+0.069°C Tb= 100.069°C      

• The given values are plugging in to equation 1 to give elevation of boiling point this was added to boiling point of (solvent) water (100°C) to give the boiling point of 0.050 m MgCl2 solution.
• The boiling point of 0.050 m MgCl2 solution is 100.069°C

To calculate the freezing point of 0.050 m MgCl2 solution

Molal freezing-point depression constant of water is 1.86°C/molal

                                  van't Hoff factor of MgCl2 is 2.7 ΔT= 2.7×1.86 °C/molal× 0.050 molal        = 0.25°C  Tf = 0.00°C-0.25°C        = -25.0°C 

• The given values are plugging in to equation 1 to give depression in freezing point this was subtract from freezing point of (solvent) water (100°C) to give the boiling point of 0.050 m MgCl2 solution