The Colligative Properties

The 4 Colligative Properties help to explain what happens to a pure solvent as solute is added to it.  

Properties of Solutions

Concept: The Colligative Properties. 

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Video Transcript

Welcome back, guys! In this new video, we’re going to take a look at the properties of solutions. Now we’re going to say that the four colligative properties help to explain what happens to a pure solvent as we add solute to it.
Let’s say that this right here on our left represents pure water, a pure solvent. And let’s say I decide to drop some NaCl in there, some regular table salt. So now we know that table salt will break up into ions. It will break up into Na+ and Cl-. They’re going to be much smaller relatively speaking, so they’re going to form – they’re going to be the solute within our solvent. So here we’re going to have solute plus solvent so we’re going to have here a solution.
Basically, the four colligative properties explains what happens to boiling point, freezing point, osmotic pressure and vapor pressure as we add solute. Basically, they explain what changes are going to happen when my pure solvent becomes a solution.

The four colligative properties of Boiling Point, Freezing Point, Vapor Pressure and Osmotic Pressure will either increase or decrease with the addition of solute to a pure solvent.  

Example: Explain what happens to each of the following properties as solute is added to a pure solvent.

a.  Boiling Point              b.  Freezing Point       

 

c. Osmotic Pressure        d. Vapor Pressure

 

 

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Example: Which of the following compounds will have the highest boiling point?

a) 0.10 m sucrose

b) 0.10 m CsBrO­­3

c) 0.35 m CH3OH

d) 0.15 m SrBr2

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Example: Pure water boils at 100oC. What is the expected boiling point of water after the addition of 13.12 g calcium bromide, CaBr2, to 325 g water. Kb = 0.512 oC/m. (MW of CaBr2 is 199.88 g/mol)

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Example: The vapor pressure of water at 100.0oC is 0.630 atm.  Determine the amount (in grams) of aluminum fluoride, AlF3, (in grams) needed to reduce its vapor pressure to 0.550 atm. (MW of AlF3 is 83.98 g/mol)

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Problem: Beta-carotene is the most important of the A vitamins. Calculate the molar mass of Beta-carotene if 25.0 mL of a solution containing 9.88 mg of Beta-carotene has an osmotic pressure of 56.16 mmHg at 30 degrees of Celsius.

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Vapor Pressure

Vapor Pressure is achieved by the equilibrium rate of vaporization equaling the rate of condensation. 

Concept: Defining Vapor Pressure. 

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Video Transcript

Hey guys! In this new video, we’re going to take a look at the liquid state and in particular, the effect it has in vapor pressure. We’re going to say that vapor pressure is defined as the partial pressure of paper molecules above the surface of the liquid under the dynamic equilibrium conditions of vaporization and condensation.
What does this mean? All this means is that vapor pressure is when the rates of condensation and vaporization are equal to one another. Think about it. In practical terms, let's say you’re boiling a pot of water and it has a lid. I know I’m not a great drawer but I'm trying here. In this pot we have liquid. What do we do to it? We're going to heat it up. Once we start heating up a pot of water, what's going to happen? Slowly but surely, the water's going to start to boil. Once it starts to boil, this liquid is going to get vaporized into gas. We’re going to actually have gas molecules exiting the liquid.
Now eventually, those gas molecules will reach to the top of the lid, which is considerably cooler than the actual boiling water. Once these gas molecules touch the lid, what's going to happen? They’re going to condense back into a liquid. What's going to happen now is they’re going to start dripping down back into the boiling water only to be vaporized once again into gas, go back up, condense again, drop back down, vaporize again, go back up. This is a continuous process.
We’re saying vapor pressure is when the rate vaporization or evaporation equals the rate of condensation. Remember, vaporization we're going from liquid to gas and then condensation, we're going from gas back down to liquid. Vapor pressure is basically when these things are in equilibrium with each other.

Example: The vapor pressure of pure liquid A is 550 torr and the vapor pressure of pure liquid B is 320 torr at room temperature. If the vapor pressure of a solution containing A and B is 465 torr, what is the mole fraction of A in the solution?

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Example: Determine the vapor pressure lowering associated with 1.32 m C6H12O6 solution (MW: 180.156 g/mol) at 25oC.

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Problem: The following boiling points belong to one of the following compounds: 117°C, 78°C, 34.5°C and 23°C.

CH3-O-CH3                                                                

CH3CH2OH                               

CH3CH2-O-CH2CH3                            

CH3CH2CH2CH2OH

a)  Which boiling point goes with what compound?

b) If each of the following substances were placed in separate sealed clear bottles at room temperature, could you identify one of the substances right away? 

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The Colligative Properties Additional Practice Problems

Using freezing point depression to find molecular weight

Mass of Lauric Acid (g)                                                                     8.003

Mass of benzoic Acid (g)                                                                   1.010

Freezing temperature of pure lauric acid (C°)                                   43.84

Freezing temperature of the benzoic acid-lauric acid mixture (C°)   39.05

a. Calculate molality (m), in mol/kg, using the formula Δt = K * m. The Kf value for lauric acid is 3.9°C•kg/mol.

 

b. Calculate moles of benzoic acid solute, using the molality and the mass (in kg) of lauric acid solvent.

 

 

c. Calculate the experimental molecular weight of benzoic acid, in g/mol.

 

 

d. Determine the accepted molecular weight of benzoic acid from its formula, C 6H5COOH.

 

e. Calculate the percent discrepancy between the experimental and accepted values.

 

 

 

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Lysozyme is an enzyme that cleaves cell walls. A 0.100 L aqueous solution of lysozyme that contains 75.0 mg of the enzyme has an osmotic pressure of 1.00 torr at 25°C. What is the molecular weight of lysozyme?

a. 1.02 x 10 3 g/mol

b. 1.39 x 10 3 g/mol

c. 1.06 x 10 5 g/mol

d. 1.39 x 10 4 g/mol

e. 1.06 x 10 7 g/mol

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A 0.25 gram sample of an unknown compound (a nonelectrolyte where i = 1) is dissolved in 225 mL of water. This solution is found to have an osmotic pressure of 342 torr at a temperature of 28°C. What is the molar mass of the unknown compound?

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When 1.744 g of non-volatile solute of unknown molar mass is dissolved in 122.2 g of pure acetic acid, the boiling point of the acetic acid is elevated by 0.246 °C. Compute the molar mass of the solution. Kb for acetic acid os 3.07 °C/m. 

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75.225 g of an unknown compound is dissolved in 455.00 g of H 2O. A scientist measures that the solution now freezes at -9.26°C. Assuming that the compound is NOT ionic, what is the molecular weight of the compound kf = 1.86°Cm-1.

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Which statements below are TRUE regarding colligative properties?

I) The magnitude depends on the quantity of the solute particles.

II) A 0.1 m CaCl 2 aqueous solution has a lower freezing point than a 0.1 m C 12 H 22 O 11 aqueous solution.

III) The vapor pressure of a solution is proportional to the mole fraction of solvent in solution.

 

A) II only

B) III only

C) both I and III

D) both II and III

E) I, II and III

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Membrane osmometry is a technique used to find the molecular weight of polymers by determining the osmotic pressure of a solvent across a semipermeable membrane. What is the molecular weight of the water soluble polymer, polyacrylamide, with a concentration of 100 g/L that produces an osmotic pressure of 8 × 10−3 atm at 25C?

1. 30,900 g/mol

2. 305,000 g/mol

3. 2,600,000 g/mol

4. 25,600 g/mol

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A solution that contains 12.6 g of a solid nonvolatile nonelectrolyte dissolved in 400 g of benzene freezes at 3.6°C. The normal freezing point of benzene is 5.5°C. What is the molar mass of the solute? (Kf for benzene is 4.96°C kg/mole)

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The osmotic pressure of a solution containing 2.46 mg of an unknown protein in 50.0 mL of solution is 1.66 atm at 25 oC? Find the molar mass of the unknown protein. 

a. 0.766 g/mol

b. 0.587 g/mol

c. 0.725 g/mol

d. 0.487 g/mol

e. 0.357 g/mol 

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Lauryl alcohol, a component of coconut oil, is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1°C. What is the molar mass of lauryl alcohol? Kf (benzene) = 5.12°C/m; T(benzene) = 5.50°C.

a) 270 g/mol

b) 13.7 g/mol

c) 62.4 g/mol

d) 46.5 g/mol

e) 183 g/mol

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Identify the colligative property.

A) vapor pressure lowering

B) freezing point depression

C) boiling point elevation

D) osmotic pressure

E) all of the above

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0.150 g of an unknown substance are added to water to make 100.00 mL of solution. The osmotic pressure of the solution is measured to be 477.688 torr at 44.0 oC. What is the molecular weight of the solute? (R= 0.08206 L atm/mol K)

A. 8.62 g/mol
B. 91.34 g/mol
C. 24.18 g/mol
D. 67.05 g/mol
E. 62.08 g/mol  

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You have a solution that contains 25 g of an unknown non-ionizing polymer in a 250 mL solution of chloroform (CHCl3). The osmotic pressure of this solution at 25C is found to be 0.056 atm. What is the molecular weight of the unknown solid? 

A. 43,500 g mol −1

B. 1,740 g mol −1

C. 10,870 g mol −1

D. 3,650 g mol −1

E. 41,000 g mol −1

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An unknown compound shows an elemental analysis of 40% C, 6.7% H and 53.3% O. When 289.418 g of the unknown compound is dissolved in 1.8 L of water, the freezing point of the solution is lowered by 3.32oC. Assuming that the compound is a nonvolatile, calculate its molar mass and its molecular formula. The freezing point constant, Kf, of water is 1.86 oC/m. 

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A solution of 2.50 g of a nonelectrolyte is dissolved in 10.0 g of water and the solution freezes at -3.720°C. What is the molecular weight of the nonelectrolyte?

Kf(H2O) = 1.860°C/m

a) 50.0 g/mol

b) 75.0 g/mol

c) 100 g/mol

d) 125 g/mol

e) 150 g/mol

 

 

 

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 A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass of the unknown compound?

A) 223 g/mol      

B) 294 g/mol      

C) 448 g/mol      

D) 195 g/mol      

E) 341 g/mol

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Which of the following is not a colligative property?

a) vapor pressure lowering

b) Henry's Law

c) freezing point depression

d) osmotic pressure

e) boiling point elevation

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Which of the following is true for all colligative properties?

a) The measurement must be made at constant temperature for all colligative properties

b) The measurement must be made at constant pressure for all colligative properties

c) The measurement must be made in a closed container

d) It is the number of particles that is important and not their type

e) The intermolecular forces are responsible for the colligative properties

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Colligative properties are similar in that they all ________

a. lower the activation energy of the solvent

b. have no effect on the properties of solution

c. describe colloids

d. are due to solvent-solvent chemical interactions

e. depend on the number of solute particles in solution

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 When a nonvolatile solute is added to a volatile solvent, the solution vapor pressure, the boiling point ___________________ the freezing point__________ and the osmotic pressure across a semi permeable membrane _________________________.

 

a) decreases, increases. decreases, increases

b) increases, increases, decreases, increases

c) increases, decreases, increases, decreases

d) decreases, decreases, increases, decreases

e) decreases; increases. decreases. decreases

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Which of the following is not a colligative property?

a. osmotic pressure

b. Henry’s Law

c. vapor pressure lowering

d. freezing point depression

e. boiling point elevation

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If a substance has weak intermolecular forces, it has a __________ boiling point, a ___________ heat of vaporization and a ______________ vapor pressure.

a) high, large, low

b) low, small, high

c) low, large, low

d) high, small, low

e) high, large, high

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