Respuesta :
Answer:
The freezing point for the solution is -8.9°C
Explanation:
We must combine the two colligative properties of freezing point depression and boiling point elevation to solve this excersise.
ΔT = Kb . m
ΔT = T° boiling point of solution - T° boiling of pure solvent
Kb = Ebullioscopic constant
m = molality
102.45°C - 100°C = 0.51°C/m . m
2.45°C / 0.51 m /°C = m → 4.80 mol/kg
We found out the molality in the boiling point elevation to replace at the freezing point depression formula
ΔT = Kf . m
ΔT = T° freezing pure solvent - T° freezing solution
Kf = Cryoscopic constant
0°C - T° freezing solution = 1.86°C/m . 4.80 m
T° freezing solution = - 8.9°C
The freezing point is the temperature at which any substance freezes. The freezing point for the solution is -8.9 degrees celsius.
What is the freezing point?
The freezing point is the property of the solution or the substance that is given by the Cryoscopic constant.
Using the boiling point elevation the molality can be calculated as:
[tex]\begin{aligned}\rm \Delta T &= \rm Kb\times m\\\\\rm \Delta T &= \rm T^{\circ} \text{boiling point of solution} - \rm T^{\circ} \text{boiling of pure solvent}\end{aligned}[/tex]
Given,
The boiling point of solution = 102.45 degrees celsius
The boiling point of pure solvent = 100 degrees celsius
Ebullioscopic constant (Kb) = 0.51 Degree celsius per m
Substituting values in the above equation:
[tex]\begin{aligned} \rm molality &= \dfrac{102.45 - 100}{0.51}\\\\&= 4.80 \;\rm mol/kg\end{aligned}[/tex]
Now, the freezing point of the solution can be calculated as:
[tex]\begin{aligned} \rm \Delta T &= \rm Kf \times m\\\\\rm T^{\circ} \text{freezing pure solvent} - \rm T^{\circ} \text{freezing solution} &= 1.86^{\circ} \;\rm C/m \times 4.80 \;\rm m\\\\\rm T^{\circ} \rm freezing \;solution &= 8.9 ^{\circ}\;\rm C\end{aligned}[/tex]
Therefore, the freezing point of the solution is -8.9 degrees celsius.
Learn more about a freezing point here:
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