Respuesta :
Answer:
T°fussion of solution is -18°C
Explanation:
We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1
First of all, we apply boiling point elevation
ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
Kb = ebuliloscopic constant
105°C - 100° = 0.512 °C kg/mol . m . 1
5°C / 0.512 °C mol/kg = m
9.7 mol/kg = m
Now that we have the molality we can apply, the Freezing point depression.
ΔT = Kf . m . i
Kf = cryoscopic constant
0° - (T°fussion of solution) = 1.86 °C/m . 9.76 m . 1
- (1.86°C /m . 9.7 m) = T°fussion of solution
- 18°C = T°fussion of solution
The boiling point of standard water is 100 degree Celsius, with the addition of solute the boiling point is elevated. The freezing point of the solution will be -18.04 degree Celsius.
What is boiling point?
The boiling point is the temperature at which the liquid is converted to vapor. The change in boiling point of the aqueous solution gives the molality of the solution as:
[tex]\rm \Delta T=ebuliloscopic constant\;\times\;molality\;\times\;von't\;hoff\;factor\\105^\circ C-100^\circ C=0.512^\circ C.kg/mol\;\times\;1\;\times\;m\\9.7\;mol/kg=m[/tex]
The depression in freezing point from molality is given as;
[tex]\rm \Delta T=K_f\;\times\;molality\;\times\;i\\\Delta\;T=1.86\;^\circ C/m\;\times\;9.7\;\times\;1\\\Delta T=18.04\;^\circ C\\[/tex]
The freezing point of aqueous water is zero degree Celsius. The freezing point of the solution will be:
[tex]\rm \Delta T=0^\circ\;C-New\;freezing\;point\\18.04^\circ\;C=0-New\;freezing\;point\\New\;freezing\;point=-18.04^\circ C[/tex]
The freezing point of the solution is -18.04 degree Celsius.
Learn more about boiling point, here:
https://brainly.com/question/2153588
