Calculate the freezing point and boiling point in each solution, assuming comple

Question

Calculate the freezing point and boiling point in each solution, assuming complete dissociation of the solute. a) Calculate the freezing point of a solution containing 12.2 g FeCl3 in 158 g water. b) Calculate the boiling point of a solution above. c) Calculate the freezing point of a solution containing 4.5 % KCl by mass (in water). d) Calculate the boiling point of a solution above. e) Calculate the freezing point of a solution containing 0.170 m MgF2. f) Calculate the boiling point of a solution above.

Explanation / Answer

we will do each part following the same steps.

1) finding molality(m)

2) calculating vant-hoff factor(i)

3)plugging in values in formula delta T= i*m* Kf/Bf

a)

molality:

moles of FeCl3= 12.2/162.2= 0.0752

m= 0.0752/0.158= 0.476 moles/Kg

FeCl3 disociates to give 4 ions:
FeCl3 Fe 3+ + 3Cl-
i=4

Kf for water = 1.86°C/m
Therefore your calculation becomes:
delta T= 4*1.86*0.476 = 3.54°C
freezing point= 0-3.54= -3.54 degree celcius

similarly

delta T= i*Bf*m

delta T= 0.512*4*0.476= 0.9748

T=100+.9748= 100.9748 degree celcius

c)

4.5 % solution means 4.5 g of KCl in 100 g water

moles= 4.5/74.55= 0.0604 moles

mass of water= 100-4.5= 95.5 g

m= 0.632 moles/Kg

KCl dissociates to form 2 ions.

so i=2

delta T= 2.35

T= -2.35 degree celcius

delta Tb= 0.647

Tb=100.647 degrees

e)

molality= 0.17

MgF2 disociates to give 3 ions
i=3

Kf for water = 1.86°C/m
Therefore your calculation becomes:
delta T= 3*1.86*0.17  = 0.9486°C
freezing point= -9486 degree celcius

similarly

delta T= i*Bf*m

delta T= 0.512*3*0.17= 0.261

T= 100.261 degree celcius

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