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 / Moles of NaCl / Moles of All Solutes / Mass of Solvent / Molality (all solutes)
B / 0.20 / 0.40 / 100 g / 4
C / 0.10 / 0.20 / 100 g / 2
D / 0.05 / 0.10 / 100 g / 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

B / C / D / E / F
Measured Freezing Point in Part D / -8°C / -8°C / 0°C / -3°C / -2°C
Freezing Point of Water in part C / 0°C / 0°C / 0°C / 0°C / 0°C
ΔT Freezing Point Depression / 8°C / 8°C / 0°C / 3°C / 2°C

Part D:

  1. 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=Tpuresolvent -Tsolution=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 C12H22O11. 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: (12*12 + 22*1 + 11*16) = 342 g/mol

Absolute % error: (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% (negativeforloss)

Indicate if the solution was hypertonic, hypotonic, or isotonic.

Solution / mass difference / % difference / tonicity
A / 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%