Colligative Properties Date
Part A:
1. Complete the following tables with correct significant figures and units. You must show calculations for
determining each quantity.
Solution
B
C
D
Moles of NaCl
0.20
0.10
0.05
Moles of All Solutes
0.40
0.20
0.10
Mass of Solvent
100 g
100 g
100 g
Molality (all solutes)
4
2
1
Solution B
The number of moles of NaCl: 11.65 g/ (58.44 g/mol) = 0.20 moles
The number of moles of all solutes (including cations and anions): 2*0.2 = 0.40 moles
Mass of solvent (assume that the density of water is 1.000 g/mL): 100 g
The molality of all solutes: 1000*(0.4 /100) = 4 m
Solution C
The number of moles of NaCl: 5.82 g / (58.44 g/mol) = 0.10 moles
The number of moles of all solutes (including cations and anions): 2*0.1 = 0.20 moles
Mass of solvent (assume that the density of water is 1.000 g/mL): 100 g
The molality of all solutes: 1000*(0.2/100) = 2 m
Colligative Properties
Solution D
The number of moles of NaCl: 2.92 / (58.44 g/mol) = 0.05 moles
The number of moles of all solutes (including cations and anions): 2*0.05 = 0.10 moles
Mass of solvent (assume that the density of water is 1.000 g/mL): 100 g
The molality of all solutes: 1000 * (0.1/100) = 1 m
Part C:
2. Using the measured freezing point of distilled water in Part C, calculate the freezing point depression ΔT for NaCl
solutions and sucrose solutions from Part D. Complete the following table. The freezing point depression ΔT is
calculated by
ΔT = (Freezing point of water in Part C) − (Measured freezing point in Part D)
The entries in the row “Freezing Point of Water in Part C” of the table should have the same value.
The entries in the row “Freezing Point Depression” of the table should be positive
Measured Freezing Point in Part D
Freezing Point of Water in part C
ΔT Freezing Point Depression
B
-8°C
0°C
8°C
C
-8°C
0°C
8°C
D
0°C
0°C
0°C
E
-3°C
0°C
3°C
F
-2°C
0°C
2°C
Part D:
2. Calculate the molal freezing point depression constant (Kf) by using the NaCl solution data. Use Equation
(3), the molality of all solutes in a NaCl solution and the freezing point depression (ΔT). Show your work
here.
(Eq. 3) ΔT=T puresolvent -T solution=Kf . m solute
Solution B: The value of Kf = ΔT / m solute = 8°C / (4 m) = 2 °C·kg/mol
Solution C: The value of Kf = ΔT / m solute = 8°C / (2 m) = 4 °C·kg/mol
Solution D: The value of Kf = ΔT / m solute = 0°C / (1 m) = 0 °C·kg/mol
4. Calculate the average of three Kf values of water from the NaCl solution data. The published value for Kf of water
is 1.86°C·kg/mol (use this as the true value). Determine the accuracy of the measurement by the absolute % error:
averagetruevalueabsolute%error100%truevalue−=×
Show your work here:
Absolute % error = Average - true value/ true value x 100%
Average: (2 + 4 + 0)/3 °C·kg/mol = 2 °C·kg/mol
Absolute % error = (2 - 1.86) / 1.86 * 100% = 7.53%
Part D:
5. Determine the molar mass of sucrose from the sucrose solution data.
a. Calculate the molality of sucrose in Solutions E and F by sing Equation (3) and the freezing point depression (ΔT)
of sucrose solution data. Assume that the value for Kf of water is 1.86°C·kg/mol.
Do not use Equation (2) to answer this question. Use Equation (3) instead. Show your work here.
Solution E: Molality of Sucrose = ΔT / Kf = 3 °C / (1.86 °C*kg/mol) = 1.61 m
Solution F: Molality of Sucrose = ΔT / Kf
= 2 °C / (1.86 °C*kg/mol) = 1.08 m
b. Calculate the number of moles of sucrose in Solutions E and F. Use the amount of water used from Part A and the
values of molality of sucrose calculated in Question 5. Assume that the density of water is 1.000 g/mL.
Solution E: Moles of sucrose = (100 g)*(1.61 mol/kg) = 0.161 moles
Solution F: Moles of sucrose = (100 g)*(1.08 mol/kg) = 0.108 moles
c. Calculate the molar mass of sucrose. Use the mass of sucrose used in Part A and the calculated values of moles of
sucrose above.
Solution E: Molar mass of sucrose = (34.24g) / 0.161 moles = 212.67 g/mol
Solution F: Molar mass of sucrose = (17.10g) / 0.108 moles = 158.33 g/mol
6. Calculate the average of molar mass of sucrose. The molecular formula of sucrose is C 12H22O11. Determine the
accuracy of the measurement by the absolute % error.
Average: (212.67 + 158.33) / 2 = 185.5 g/mol
True value of molar mass of sucrose:
Absolute % error:
(12*12 + 22*1 + 11*16) = 342 g/mol
(342-185.5)/342 * 100% = 45.76%
Part E:
7. Complete the table: Calculate the difference in the mass of the celery stalk for each solution. Also, calculate the
percentage difference by
mass difference= mass(after)- mass(initial)
(negative for loss)
% difference = mass difference/ mass initial x 100% (negative for loss)
Indicate if the solution was hypertonic, hypotonic, or isotonic.
Solution
% difference
tonicity
A
mass
difference
0.79
10%
Hypotonic
B
-0.74
-12%
Hypertonic
C
-0.66
-9%
Hypertonic
D
-0.41
-6%
Hypertonic
E
-0.39
-6%
Hypertonic
F
0.01
0%
Isotonic
Show you work here for calculating mass difference and % difference:
Solution A: Mass difference = 8.53 - 7.74 = 0.79 and % difference = 0.79 / 7.74 * 100% = 10%
Solution B: Mass difference = 5.58 - 6.32 = -0.74 and % difference = -0.74 / 6.32 * 100% = -12%
Solution C: Mass difference = 6.58 - 7.24 = -0.66 and % difference = -0.66/7.24 *100% = -9%
Solution D: Mass difference = 5.99 - 6.40 = -0.41 and % difference = -0.41/6.40*100% = -6%
Solution E: Mass difference = 6.51- 6.90 = -0.39 and % difference = -0.39/6.90*100% = -6%
Solution F: Mass difference = 7.40 - 7.39 = 0.01and % difference = 0.01/7.39*100% = 0%