Honors Chemistry
Name ____________________________
Colligative Properties Worksheet II
Date _____/_____/_____
Period _____
1. What is a Colligative property?
Properties that depend on the concentration of solute particles but not on their
identity. (For example: Freezing-point Depression, Boiling-point Elevation, and
Osmotic Pressure)
2. What are freezing point depression, boiling point elevation, and osmotic pressure? List the
formulas that are used to calculate each.
Freezing Point Depression (Tf) – is the difference between the freezing points of the
pure solvent and a solution of a nonelectrolyte in the
solvent, and it is directly proportional to the molal
concentration of the solution.
For an electrolyte solution, the number of ions (i) must
be taken into account. (Tf = i Kf m)
Boiling-point Elevation (Tb) – is the difference between the boiling points of the pure
solvent and a nonelectrolyte solution of that solvent, and it
is directly proportional to the molal concentration of the
solution.
For an electrolyte solution, the number of ions (i) must be
taken unto account. (Tb = i kb m)
Osmotic Pressure – is the external pressure that must be applied to stop osmosis. The
number of ions (i) must also be taken into account in osmotic
pressure calculations.
 = iMRT
3. What is the molal freezing-point constant (Kf) for water?
Kf = 1.86 oC ▪ kg
mole
4. What is the molal boiling point constant (Kb) for water?
Kb = 0.52 °C
m
5. What is the molality of a solution of 1.25 moles of sugar dissolved in 0.750 kg of water?
What is the boiling point and freezing point of the resulting solution?
a.
m = 1.25 mole/0.750 kg = 1.67 m

b.
Tb = i kb m
Tb = (1)(0.52 °C/m)( 1.67 m)
Tb = 0.87 °C
BP = 100.00 °C + 0.87 °C = 100.87 °C
c.
Tf = i kf m
Tf = (1)(1.86 oC ▪ kg/mole)( 1.67 m)
Tf = 3.11 oC
FP = 0.000 oC – 3.11 oC = – 3.11 oC
6. An aqueous solution contains 3.00 % phenylalanine (C9H11NO2) (MM = 165.21 g/mole) by
mass at 25.0 oC. Assume the phenylalanine is non-ionic and volatile and that the density of
the solution is 1.00 g/mL. Calculate the following:
m = (3.00 g)/(165.21 g/mole)/(0.097 kg) = 0.19 m
a. The freezing point of the solution (Tf)
Tf = i kfm
Tf = (1)(1.86 oC ▪ kg/mole)(0.19 m)
Tf = 0.35 oC
FP = – 0.35 oC
b. The boiling point of the solution (Tb)
Tb = i kbm
Tb = (1)(0.52 °C/m)(0.19 m)
Tb = 0.099 °C
BP = 100.000 °C + 0.099 °C = 100.099 °C
c. The osmotic pressure of the solution ()
M = (3.00 g)/(165.21 g/mole)/0.100 L = 0.182 M
 = iMRT
 = (1) (0.182 M) (0.0821 L ▪ atm / mole ▪ K) (298 K)
 = 4.45 atm
7. An aqueous solution was found to have an initial freezing point of -2.15 oC. What was the
molality of the solution?
Tf = i kfm
2.15 oC = (1)(1.86 oC ▪ kg/mole) m
m = 1.16 m
8. A solution of glucose (MM = 180.18 g/mole) in 500. g of water had an initial boiling point of
102.25 oC.
a. What is the molality of the solution?
Tb = i kbm
2.25 oC = (1)(0.52 °C/m) m
m = 4.3 m
b. How many moles of glucose were added?
m = moles of solute / Kg of solvent
4.3 m = moles of glucose / 0.500 Kg
moles of glucose = 2.2 moles
c. How many grams of glucose were added?
grams of glucose = (2.2 moles of glucose)(180.18 g/mole)
–
grams of glucose = 400 g
9. Pure toluene (C7H8) has a normal boiling point of 110.60 oC. A solution of 7.80 g of
anthracene (C14H10), in 100.0 g of toluene has a boiling point of 112.06 oC.
a. What is the molality of the solution?
m = 7.80 g of C14H10 / 178.24 g/mole = 0.0438 mole / 0.100 Kg = 0.438 m
b. What is the molal boiling point elevation constant for toluene?
1.46 oC = (1) Kb (0.438 m)
Kb = 3.33 oC/m
10. What is the osmotic pressure (in atmospheres) of a 12.36 M aqueous solution of urea at
22 C?
 = iMRT
 = (1) (12.36 M) (0.0821 L ▪ atm / mole ▪ K) (295 K)
 = 299 atm
11. What is the molarity of a solution in which the osmotic pressure is 80.7 atm at 27 oC?
Assume the solute is a nonelectrolyte.
 = iMRT
M =  / iRT = (80.7 atm) / (1)( 0.0821 L ▪ atm / mole ▪ K) (300. K)
Molarity = 3.28 M
12. At what temperature, in Celsius, would the osmotic pressure have to be 347.5 kPa in order
to stop osmosis if there was 24.37 g of glucose (MM = 180.18 g/mole) dissolve in enough
water to make a 1.00 L solution?
a. Calculate moles of glucose:
24.37 g / 180.18 g/mole = 0.1353 mole
b. Calculate molarity of solution:
M = 0.1353 mole / 1.00 L = 0.135 M
c. Calculate the osmotic pressure in atm:
= 347.5 kPa / 101.3 kPa = 3.43 atm
d. Calculate the temperature:
T =  / iMR = (3.43 atm) / (1) (0.135 M) (0.0821 L ▪ atm / mole ▪ K)
T = 309 K = 36 oC
13. Seawater is about 3.5% (by weight) dissolved solids, almost all of which is NaCl. Calculate
the normal boiling point of seawater.
a. Calculate moles of NaCl:
3.5 g / 58.5 g/mole = 0.060 mole
b. Calculate molality of NaCl:
m = 0.060 mole / 0.0965 kg = 0.62 m
c. Use boiling point elevation constant:
T = i Kb m
T = (2) (0.52 °C/m) (0.62 m) = 0.64 °C
The seawater boils at 100.64 °C
14. Calculate the freezing point of a solution of 5.00 g of diphenyl C12H10 and 7.50 g of
naphthalene, C10H8 dissolved in 200.0 g of benzene (fp = 5.5 °C and Kf = 5.12 °C/m).
There is a tiny curve in this problem, but keep in mind that Colligative properties are all
about how many particles in solution and nothing else.
a. So, the key to this problem is to calculate moles of each substance and add them
together:
(5.00 g / 154.2 g/mole) + (7.50 g / 128.2 g/mole) = 0.0909 mole
b. Now, we calculate the molality:
0.0909 mole / 0.200 kg = 0.455 m
c. The last step is to calculate the freezing point depression:

T = i Kf m
T = (1) (5.12 °C/m) (0.455 m) = 2.33 °C
The solution freezes at 5.5 – 2.33 = 3.2 °C
15. The freezing point depression of a 0.10 m solution of HF(aq) solution is -0.201 °C.
Calculate the percent dissociation of HF(aq).
a. The freezing point depression equation is:
T = i Kf m
The only value we do not know is i:
0.201 °C = i(1.86 °C/m) (0.10m)
i = 1.1
The HF is about 10. % dissociated.
16. 2.60 grams of a compound know to contain only indium and chlorine is dissolved in 50.0 g
of tin (IV) chloride (Kb = 9.43 °C kg mol¯1). The normal boiling point is raised from 114.1 °C
for pure SnCl4 to 116.3 °C for the solution. What is the molecular weight and probably
molecular formula for the solute?
Solution
Keep in mind that the units on molecular weight are grams per mole. We already have the
grams, what we now need are the moles that 2.60 grams represent. We can get that by
the molality (moles solute / kg solvent):
T = i Kb m
2.2 = 9.43 (x / 0.050); x = 0.01168 mol
Now we calculate the molecular weight:
2.60 g / 0.01168 mol = 222.6 g/mol
The probable molecular formula is InCl3, but I will let you ponder on how I did it.
17. 0.64 g of adrenaline in 36.0 g of CCl4 (Kb = 5.03 °C / m) produces a boiling point elevation
of 0.49 °C. What is adrenaline's molecular weight.
a. T = i Kb m
0.49 °C = (1)(5.03 °C / m)(moles of solute/ 0.0360 kg)
moles of solute = 0.0035 moles
b. From here it is a simple step: grams divided by moles equals the desired answer.
0.64 g / 0.0035 mole = 180 g/mole
18. Salt is often used to remove ice from roads and sidewalks. Explain how this process works
in terms of Colligative properties.
When the salt dissolves into solution it ionizes. It is the production of ions that
causes the resulting solution to have a lower freezing point than the solvent
(water/snow ice). The more ions that are produced in solution the lower the
resulting freezing point.
19. Which salt, NaCl or CaCl2, has a greater effect on freezing point? Explain.
Since CaCl2 produces more ions in solution, it will cause the resulting solution to
have a lower freezing point. The more particles dissolved in the solvent the lower
the freezing point.