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G E N E R A L P R O P E R T I E S O F S O L U T I O N S. 1. A solution is a homogeneous mixture of two or more components. 2. It has variable composition. 3. The dissolved solute is molecular or ionic in size. 4. A solution may be either colored or colorless nut is generally transparent.

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g e n e r a l p r o p e r t i e s o f s o l u t i o n s
GENERAL PROPERTIES OF SOLUTIONS

1. A solution is a homogeneous mixture of two or more components.

2. It has variable composition.

3. The dissolved solute is molecular or ionic in size.

4. A solution may be either colored or colorless nut is generally transparent.

5. The solute remains uniformly distributed throughout the solution and will not settle out through time.

6. The solute can be separated from the solvent by physical methods.

colligative properties of solutions
Colligative Properties of Solutions
  • Depends on the concentration of the solute particles and not on the identity of the solute.
  • Dissolved particles alter and interfere with the dynamic process of a solution.
  • NOTE: DT=Tf-Tior in this caseDT=Tsolution-Tsolvent
  • Boiling point elevation
  • Freezing point depression
  • Osmosis
  • Vapor pressure lowering
colligative properties d t k m
Colligative PropertiesDT = k m

1. Calculate the molality of a solution consisting of 5.34 g of aluminum sulfate in 200.0 mL of pure water at 24oC?

2. The boiling point of a solution containing 5.55 g of an unknown nonvolatile substance dissolved in 15.0 g of water is 103.3 oC. Calculate the molar mass of the compound.

3. Calculate the freezing point of a 500.0 mL sample of aluminum sulfate solution (kf = 1.86 oC kg / mol) containing 16.0 g of solute.

molality
MOLALITY
  • Molality = moles of solute per kg of solvent
  • m = nsolute / kg solvent
  • If the concentration of a solution is given in terms of molality, it is referred to as a molal solution.

Q. Calculate the molality of a solution consisting of 25 g of KCl in 250.0 mL of pure water at 20oC?

First calculate the mass in kilograms of solvent using the density of solvent:

250.0 mL of H2O (1 g/ 1 mL) = 250.0 g of H2O (1 kg / 1000 g) = 0.2500 kg of H2O

Next calculate the moles of solute using the molar mass:

25 g KCl (1 mol / 54.5 g) = 0.46 moles of solute

Lastly calculate the molality:

m = n / kg = 0.46 mol / 0.2500 kg = 1.8 m (molal) solution

freezing point depression d t f k f m
Freezing Point DepressionDTf = - kfm

Q. Estimate the freezing point of a 2.00 L sample of seawater (kf = 1.86 oC kg / mol), which has the following composition:

0.458 mol of Na+ 0.052 mol of Mg2+ 0.010 mol Ca2+

0.010 mol K+ 0.533 mol Cl- 0.002 mol HCO3-

0.001 mol Br- 0.001 mol neutral species.

Since colligative properties are dependent on the NUMBER of particles and not the character of the particles, you must first add up all the moles of solute in the solution.

Total moles = 1.067 moles of solute

Now calculate the molality of the solution:

m = moles of solute / kg of solvent = 1.067 mol / 2.00 kg

= 0.5335 mol/kg

Last calculate the temperature change:

DTf = - kfm = -(1.86 oC kg/mol) (0.5335 mol/kg) = 0.992 oC

The freezing point of seawater is Tsolvent - DT = 0 oC - 0.992 oC

= - 0.992 oC

boiling point elevation d t b k b m
Boiling Point ElevationDTb = kbm

Q. The boiling point of a solution containing 40.0 g of an unknown substance dissolved in 100.0 g of water is 105.3 oC . Calculate the molar mass of the compound.

Since the solvent is water, the change in temperature (DT) would be 105.3 - 100.0 oC = 5.3 oC. You can also find the kb in the table in your textbook, kb = 0.512 oC kg/mol.

From this data, you can calculate the molality:

m = DTb / kb = 5.3 oC / 0.512 oC kg/mol = 10.4 mol/kg

Molality is also defined as the moles of solute per kg of solvent:

m = n /(kg solvent), can be rearranged to be n = m (kg of solvent)

n = 10.4 mol /kg (0.1000 kg) = 1.04 mol of solute

The molar mass can be calculated by using the equation, MW = m/n

MW = 40.0 g / 1.04 mol= 38.5 g/mol

practice problems cp
PRACTICE PROBLEMS # CP

Short essay: (answers on next slide)

1. Excluding any possible chemical reactions, which would be more effective as an antifreeze; a solution containing 25 m methyl alcohol (CH3OH) or 25 m KCl?

Application:

2. What is the freezing point of an aqueous sugar (C12H22O11) solution that boils at 110oC?

3. When 256 g of a nonvolatile, nonelectrolyte unknown were dissolved in 499 g of water, the freezing point was found to be –2.79oC. The molar mass of the unknown solute is?

a) 357 b) 62.0 c) 768 d) 342

-36.3oC

D

slide8
Short essay answers:

1. Excluding any secondary chemical reactions, which would be more effective as an antifreeze; a solution containing 25 m methyl alcohol (CH3OH) or 25 m KCl?

KCl because colligative properties depend on the number of solute particles and not on the nature of the particles therefore you must calculate the amount of solute particles in each case.

Proof:

If you assume you have 1 kg of solution, a 25 molal (moles of solute per kg of solvent) solution of methyl alcohol would contain 25 moles of solute. In one 1 kg of a 25 molal KCl solution there would be 25 moles of K+ ions and 25 moles of Cl- ions (because KCl completely dissociates in water) therefore KCl has twice as many particles as methyl alcohol and thus be more effective.

group study problems cp
GROUP STUDY PROBLEMS # CP

Short Essay (write answers on back)

1. Which would be more effective as an antifreeze; a solution containing 25 m methyl alcohol (CH3OH) or 25 m ethyl alcohol (C2H5OH)?

Application:

___2. An aqueous solution freezes at -7.98 oC. What is its boiling temperature?

___ 3. When 42.8 g of a nonvolatile, nonelectrolyte unknown were dissolved in 423 g of water, the freezing point was found to be –2.29oC. The molar mass of the unknown solute is?