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Home , Colligative properties, Freezing, Molality, Molar mass

8.1HW .doc Colligative Properties of

Use the Equations given in your notes to solve the Colligative Property Questions.

ΔTb = m × Kb, ΔTf = m × Kf

Freezing Boiling K K Formula Point f b Point (°C) (°C/m) (°C/m) (°C)

Water H2O 0.000 100.000 1.858 0.521

Acetic acid HC2H3O2 16.60 118.5 3.59 3.08

Benzene C6H6 5.455 80.2 5.065 2.61

Camphor C10H16O 179.5 ... 40 ...

Carbon disulfide CS2 ... 46.3 ... 2.40

Cyclohexane C6H12 6.55 80.74 20.0 2.79

Ethanol C2H5OH ... 78.3 ... 1.07

1. Which solvent’s freezing point is depressed the most by the addition of a solute?

This is determined by the Freezing Point Depression constant, Kf. The substance with the highest value for Kf will be affected the most. This would be with a constant of 40.

2. Which solvent’s freezing point is depressed the least by the addition of a solute?

By the same logic as above, the substance with the lowest value for Kf will be affected the least. This is . Certainly the case could be made that disulfide and are affected the least as they do not have a constant.

3. Which solvent’s is elevated the least by the addition of a solute?

Water

4. Which solvent’s boiling point is elevated the most by the addition of a solute?

Acetic Acid

5. How does Kf relate to Kb? Kf > Kb (fill in the blank) The freezing point constant is always greater.

6. A of 58.5 grams of NaCl in 1,000 grams of water is made. At what will the solution freeze?

Calculate the Moles of in the Solution

Molar of NaCl = 58.5 (you should know how to find this by now)

58.5g _ NaCl 1_  1_ mole This gives the # of moles in the solution. 1 58.5_ grams

Calculate the of the Solution

# _ of _ moles _ of _ solute 1_ mole   1 molal solution kg _ of _ solution 1_ kg

List Variables

m = 1 molal Kf = (from chart above), 1.858 ΔTf = change in freezing temp

Write the Equation: ΔTf = m × Kf

Plug in the Appropriate Values

ΔTf = m × Kf = (1 m)(1.858 C/m) = 1.858 C (this is the change in the temperature not the temp)

Calculate the Temperature

Tf = 0 - 1.858 = -1.858 degrees Celcius

7. A solution of 146 grams of NaCl in 1,000 grams of water is made. At what temperature will the solution freeze?

Calculate the Moles of Salt in the Solution

Molar Mass of NaCl = 58.5 (you should know how to find this by now)

146g _ NaCl 1_ mole   2.5_ mole This gives the # of moles in the solution. 1 58.5_ grams Calculate the molality of the Solution

# _ of _ moles _ of _ solute 2.5_ mole   2.5 molal solution kg _ of _ solution 1_ kg

List Variables

m = 2.5 molal Kf = (from chart above), 1.858 ΔTf = change in freezing temp

Write the Equation ΔTf = m × Kf

Plug in the Appropriate Values

ΔTf = m × Kf = (1 m)(1.858 C/m) = 4.6 C (this is the change in the temperature not the temp)

Calculate the Temperature

Tf = 0 – 4.6 = -4.6 degrees Celcius

8. At what temperature will the solution described in number 6 boil? List Variables m = 1 molal Kb = (from chart above), .521 ΔTb = change in freezing temp

Write the Equation: ΔTb = m × Kb

Plug in the Appropriate Values

ΔTb = m × Kb = (1 m)(.521 C/m) = .521 C (this is the change in the temperature not the temp)

Calculate the Temperature

Tb = 100 + .521 = 100.521 degrees Celcius

9. 1 of a water solution has an unknown amount of salt in it. To find quantity of salt the sample is cooled until it freezes. It is found to freeze at -8ºC. Show all the calculations to find the number of moles in the solution. (You will need to work backwards) List Variables

m = ?, # of moles = ? Kf = (from chart above), 1.858 ΔTf = -8ºC

Write the Equation ΔTf = m × Kf

∆Tf Solve the Equation for m: m = Kf

Plug in the Appropriate Values to Determine Molality

∆Tf 8° 퐶 m = = = 4.3 푚표푙푎푙 푠표푙푢푡푖표푛 Kf 1.858 푚

Since this is in one kilogram of solvent it is 4.3 moles of solute.

10. At what temperature will a 1 molal solution of salt in Acetic Acid (Vinegar) boil?

List Variables m = 1 molal Kb = (from chart above), .3.08 ΔTb = change in freezing temp

Write the Equation: ΔTb = m × Kb

Plug in the Appropriate Values

ΔTb = m × Kb = (1 m)(3.08 C/m) = 3.08 C (this is the change in the temperature not the temp)

Calculate the Temperature

Tb = 118.5 + 3.08 = 121.58 degrees Celcius 11. At what temperature will a 2.5 molal solution of salt in Acetic Acid (Vinegar) boil?

List Variables m = 2.5 molal Kb = (from chart above), .3.08 ΔTb = change in freezing temp

Write the Equation: ΔTb = m × Kb

Plug in the Appropriate Values

ΔTb = m × Kb = (2.5 m)(3.08 C/m) = 7.7 C (this is the change in the temperature not the temp)

Calculate the Temperature

Tb = 118.5 + 7.7 = 126.2 degrees Celcius

12. A sample is given to a student in order to find its Kb . The student makes various and heats each to measure its boiling point. The solvent itself (with no solute boils at 150 ºC) Using the data below, calculate its Kb. Molarity of Boiling Solution Temp (Cº) 1 158.8 2.5 172 3 176.4

The slope or the graph represents the constant K,

푅푖푠푒 176 퐶 퐶 푆푙표푝푒 = = = 58.7 푅푢푛 3 푚표푙푎푙 푚

13. A chief ingredient of antifreeze (for your car) is liquid glycol, C2H4(OH)2 (MM = 62 g/mole). Assume C2H4(OH)2 is added to a car radiator which holds 5 liters of water.

A. How many moles should be added to 5 liters of water to lower the freezing point from 0 C to -18 C?

List Variables

m = ?, # of moles = ? Kf = (from chart above), 1.858 ΔTf = -18ºC

Write the Equation ΔTf = m × Kf

∆Tf Solve the Equation for m: m = Kf

Plug in the Appropriate Values to Determine Molality

∆T 18° 퐶 m = f = = 9.7 푚표푙푎푙 푠표푙푢푡푖표푛 Kf 1.858 푚

Use the molal Equation to determine total Moles

moles 9.7 mole X molality = = = , so x = 4.8 moles of C2H4(OH)2 kg of solvent 1 kg 5 kg

B. How many grams is this?

4.8 푚표푙푒푠 퐶 퐻 (푂퐻) 62 푔 퐶 퐻 (푂퐻) 2 4 2 × 2 4 2 = 298 푔푟푎푚푠 1 1 푚표푙푒 퐶2퐻4(푂퐻)2

C. If C2H4(OH)2 has a of 1.1 kg/liter, how many liters is this?

298 푔 퐶 퐻 (푂퐻) 1 푘푔 1 푙푖푡푒푟 2 4 2 × × = .27 푙푖푡푒푟푠 표푓 퐶 퐻 (푂퐻) 1 1,000 푔 1.1 푘푔 2 4 2

D. What happens to the of water in the radiator as you add the antifreeze?

The radiator and cooling system in the car is a fixed volume so when the antifreeze is added an equal volume of water is removed. This acts to increase the molality of the solution. This could be factored in but it would require a second equation be created.