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.

Answer 1

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