Respuesta :
Answer:
Solutions from highest to lowest freezing point:
0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol
Explanation:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f[/tex] = Depression in freezing point
i = van'T Hoff fcator
[tex]K_f[/tex] = Molal depression constant of solvent
m = molality of the solution
Higher the value of depression in freezing point at lower will be freezing temperature the solution.
1. 0.040 m glycerin
Molal depression constant of water = [tex]K_f=1.86^oC/m[/tex]
i = 1 ( organic molecule)
m = 0.040 m
[tex]\Delta T_{f,1}=1\times\times 1.86^oC/m\times 0.040 m[/tex]
[tex]\Delta T_{f,1}=0.0744^oC[/tex]
2. 0.020 m potassium bromide
Molal depression constant of water = [tex]K_f=1.86^oC/m[/tex]
i = 2 (ionic)
m = 0.020 m
[tex]\Delta T_{f,2}=2\times\times 1.86^oC/m\times 0.020 m[/tex]
[tex]\Delta T_{f,2}=0.0744^oC[/tex]
3. 0.030 m phenol
Molal depression constant of water = [tex]K_f=1.86^oC/m[/tex]
i = 1 (organic)
m = 0.030 m
[tex]\Delta T_{f,3}=1\times\times 1.86^oC/m\times 0.030 m[/tex]
[tex]\Delta T_{f,3}=0.0558^oC[/tex]
[tex]0.0744^oC=0.0744^oC > 0.0558^oC[/tex]
[tex]\Delta T_{f,1}=\Delta T_{f,2}>\Delta T_{f,3}[/tex]
Solutions from highest to lowest freezing point:
0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol
The freezing point of the water has been zero degrees Celsius. The decreasing order of the freezing point of the solutions is Glycerin = Potassium Bromide > Phenol.
What is the freezing point?
The freezing point for the solution is the temperature at which the liquid is converted to solid. With the addition of the solute, the depression in the freezing point of water is given as;
[tex]\rm \Delta T=K_f\;\times\;molality\;\times\;von't\;hoff\;factor[/tex]
- The depression in freezing point with 0.040 m Glycerin is:
[tex]\rm K_f=1.86\;^\circ C/m[/tex]
For the covalent compound, i = 1
[tex]\Delta T=1.86\;\times\;0.040\;\times\;1\\\Delta T=0.0744^\circ\;C[/tex]
The depression in freezing with glycerin is 0.0744 degrees Celsius.
- The depression in freezing with 0.020 m KBr is:
For the ionic compound, i = 2
[tex]\Delta T=1.86\;\times\;0.020\;\times\;2\\\Delta T=0.0744^\circ\;C[/tex]
The depression in freezing with KBr is 0.0744 degrees Celsius.
- The depression in freezing with 0.020 m Phenol is:
The von't Hoff factor for the organic compound is, i = 1
[tex]\Delta T=1.86\;\times\;0.030\;\times\;1\\\Delta T=0.0558^\circ\;\rm C[/tex]
The depression in freezing with Phenol is 0.0558 degrees Celsius.
Thus, the decreasing order of the freezing point is, Glycerin = Potassium Bromide > Phenol.
Learn more about freezing point, here:
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