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1 Physical Properties of Solutions Chapter 12 (semester 1/2015) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

1 Physical Properties of Solutions Chapter 12 (semester 1/2015) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display

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Page 1: 1 Physical Properties of Solutions Chapter 12 (semester 1/2015) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display

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Physical Properties of SolutionsChapter 12 (semester 1/2015)

Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

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12.1 Types of solution

A solution is a homogenous mixture of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)

The solvent is the substance present in the larger amount

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A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.

An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.

A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

Sodium acetate crystals rapidly form when a seed crystal isadded to a supersaturated solution of sodium acetate.

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12.2 Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

Percent by Mass

% by mass = x 100%mass of solutemass of solute + mass of solvent

= x 100%mass of solutemass of solution

Mole Fraction (X)

XA = moles of A

sum of moles of all components

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Example 12.2

A sample of 0.892 g of potassium chloride (KCl) is dissolved in 54.6 g of water.

What is the percent by mass of KCl in the solution?

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Concentration Units Continued

M =moles of solute

liters of solution

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

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12.4 Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

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Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities.

Suppose you have 90 g KNO3 contaminated with 10 g NaCl.

Fractional crystallization:

1. Dissolve sample in 100 mL of water at 600C

2. Cool solution to 00C

3. All NaCl will stay in solution (s = 34.2g/100g)

4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

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Temperature and Solubility

O2 gas solubility and temperature

solubility usually decreases with

increasing temperature

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12.5 Pressure and Solubility of Gases

The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

c = kP

c is the concentration (M) of the dissolved gas

P is the pressure of the gas over the solution

k is a constant for each gas (mol/L•atm) that depends only on temperature

low P

low c

high P

high c

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Chemistry In Action: The Killer Lake

Lake Nyos, West Africa

8/21/86CO2 Cloud Released

1700 Casualties

Trigger?

• earthquake

• landslide

• strong Winds

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12.6Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering

Raoult’s law

If the solution contains only one solute:

X1 = 1 – X2

P 10 - P1 = P = X2 P 1

0

P 10 = vapor pressure of pure solvent

X1 = mole fraction of the solvent

X2 = mole fraction of the solute

P1 = X1 P 10

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PA = XA P A0

PB = XB P B0

PT = PA + PB

PT = XA P A0 + XB P B

0

Ideal Solution

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Boiling-Point Elevation

Tb = Tb – T b0

Tb > T b0 Tb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

Tb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m) for a given solvent

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Freezing-Point Depression

Tf = T f – Tf0

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m) for a given solvent

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Osmotic Pressure ()

Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure () is the pressure required to stop osmosis.

dilutemore

concentrated

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HighP

LowP

Osmotic Pressure ()

= MRT

M is the molarity of the solution

R is the gas constant

T is the temperature (in K)

solvent solution

time

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Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering P1 = X1 P 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

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Example 12.10

Our first step is to calculate the molality of the solution. From Equation (12.7) and Table 12.2 we write

Because there is 0.205 mole of the solute in 1 kg of solvent, the number of moles of solute in 301 g, or 0.301 kg, of solvent is

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Example 12.11

A solution is prepared by dissolving 35.0 g of hemoglobin (Hb) in enough water to make up 1 L in volume.

If the osmotic pressure of the solution is found to be 10.0 mmHg at 25°C, calculate the molar mass of hemoglobin.