Calculate the freezing point and boiling point of each of the following solutions
A. 2.75m NaOH in water
B. 0.586m of water in ethanol
c. 1.26m of naphthalene (C10H8) in benzene
See response to earlier post.
B) 103.8 deg c
To calculate the freezing point and boiling point of a solution, we need to use the concept of colligative properties. These properties depend on the number of solute particles present in the solution, rather than the specific identity of the solute.
The freezing point depression and boiling point elevation formulas are as follows:
Freezing Point Depression (∆Tf) = Kf * m
Boiling Point Elevation (∆Tb) = Kb * m
Where:
Kf = cryoscopic constant (freezing point depression constant)
Kb = ebullioscopic constant (boiling point elevation constant)
m = molality of the solution (moles of solute per kilogram of solvent)
Now, let's calculate the freezing point and boiling point for each solution:
A. 2.75m NaOH in water:
Since we know the molality of the NaOH solution, we can calculate the freezing point depression and boiling point elevation.
For water, the cryoscopic constant (Kf) is 1.86 °C/m and the ebullioscopic constant (Kb) is 0.512 °C/m.
Freezing Point Depression (∆Tf) = Kf * m
∆Tf = 1.86 °C/m * 2.75 m
∆Tf = 5.115 °C
Therefore, the freezing point of the solution will be 5.115 °C below the freezing point of pure water.
Boiling Point Elevation (∆Tb) = Kb * m
∆Tb = 0.512 °C/m * 2.75 m
∆Tb = 1.408 °C
Therefore, the boiling point of the solution will be 1.408 °C above the boiling point of pure water.
B. 0.586m of water in ethanol:
Using the same formulas as above, we can calculate the freezing point depression and boiling point elevation for this solution.
For ethanol, the cryoscopic constant (Kf) is 1.99 °C/m and the ebullioscopic constant (Kb) is 1.22 °C/m.
Freezing Point Depression (∆Tf) = Kf * m
∆Tf = 1.99 °C/m * 0.586 m
∆Tf = 1.164 °C
Therefore, the freezing point of the solution will be 1.164 °C below the freezing point of pure ethanol.
Boiling Point Elevation (∆Tb) = Kb * m
∆Tb = 1.22 °C/m * 0.586 m
∆Tb = 0.714 °C
Therefore, the boiling point of the solution will be 0.714 °C above the boiling point of pure ethanol.
C. 1.26m of naphthalene (C10H8) in benzene:
For this solution, we need to consider the molality of naphthalene (C10H8) in benzene.
Since there are no specific values given for the cryoscopic constant (Kf) and the ebullioscopic constant (Kb) for naphthalene in benzene, we cannot directly calculate the freezing point depression and boiling point elevation. However, we can approximate them using average values.
For benzene, the average cryoscopic constant (Kf) is around 5.12 °C/m, and the average ebullioscopic constant (Kb) is around 2.53 °C/m.
Freezing Point Depression (∆Tf) = Kf * m
∆Tf = 5.12 °C/m * 1.26 m
∆Tf ≈ 6.4512 °C
Therefore, the freezing point of the solution will be approximately 6.4512 °C below the freezing point of pure benzene.
Boiling Point Elevation (∆Tb) = Kb * m
∆Tb = 2.53 °C/m * 1.26 m
∆Tb ≈ 3.1878 °C
Therefore, the boiling point of the solution will be approximately 3.1878 °C above the boiling point of pure benzene.
Please note that the calculated values are approximate, as actual cryoscopic and ebullioscopic constants may vary depending on experimental conditions and the specific properties of the solvent and solute.