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สมบัติของสารละลาย ( Colligative properties). ว 30231 ปริมาณสัมพันธ์ สถานะของสาร และเคมีไฟฟ้า. นายศราวุทธ แสงอุไร ครูวิชาการสาขาเคมี โรงเรียนมหิดลวิทยานุสรณ์. วันที่ 13 พฤศจิกายน 2552. สมบัติของสารละลาย ( Colligative properties).

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colligative properties

สมบัติของสารละลาย(Colligative properties)

ว30231 ปริมาณสัมพันธ์ สถานะของสาร และเคมีไฟฟ้า

นายศราวุทธ แสงอุไร

ครูวิชาการสาขาเคมี โรงเรียนมหิดลวิทยานุสรณ์

วันที่ 13 พฤศจิกายน 2552

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide2

สมบัติของสารละลาย (Colligative properties)

  • Vapor and osmotic pressures, bp, and mp are colligative properties
    • Depend on relative of solute and solvent particles

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

for examples
For Examples:

Vapor Pressure Reduction

Related to boiling point

Freezing Point Depression

Salt on the road

Anti-freeze in your radiator

Boiling Point Elevation

Anti-freeze in your radiator

Osmotic Pressure

Membrane diffusion

The Great Sugar Fountain!

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

vapor pressure
Vapor Pressure

Remember:

Equilibrium vapor pressure

Pressure of vapor when liq and vapor in equilibrium at specific temp

Vapor pressure of soln lower than pure solvent vapor pressure

Vapor pressure of solvent  relative # of solvent molecules in soln

i.e., solvent vapor pressure  solvent mole fraction

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

vapor pressure1
Vapor Pressure

At equilibrium, the rate of evaporation (liquid to gas) equals the rate of condensation (gas to liquid). The amount of gas is the “vapor pressure”

P

T = K

P = atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

vapor pressure2
Vapor Pressure

At equilibrium, the rate of evaporation (liquid to gas) equals the rate of condensation (gas to liquid). The amount of gas is the “vapor pressure”

P

T = K

P = atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

vapor pressure3
Vapor Pressure

At equilibrium, the rate of evaporation (liquid to gas) equals the rate of condensation (gas to liquid). The amount of gas is the “vapor pressure”

P

T = K

P = atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

raoult s law
Raoult’s Law

Psolution = Xsolvent P°solvent

So if 75% of molecules in soln are solvent molecules (0.75 = Xsolvent)

Vapor pressure of solvent (Psolvent) = 75% of P°solvent

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

colligative properties of nonelectrolyte solutions

= vapor pressure of pure solvent

0

P1 = X1 P 1

0

0

0

P 1

P 1

P 1

- P1 = DP = X2

Colligative Properties of Nonelectrolyte Solutions

X1= mole fraction of the solvent

Raoult’s law

If the solution contains only one solute:

X2= mole fraction of the solute

X1 = 1 – X2

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide10

0

0

0

0

PA = XA P A

PB = XB P B

PT = XA P A

+XB P B

Ideal Solution

PT = PA + PB

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide11

Force

A-A

Force

A-A

Force

B-B

Force

B-B

Force

A-B

Force

A-B

<

>

&

&

PT is greater than

predicted by Raoults’s law

PT is less than

predicted by Raoults’s law

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

problem
Problem

The vapor pressure of pure acetone (CH3COCH3) at 30°C is 0.3270 atm. Suppose 15.0 g of benzophenone, C13H10O (MW = 182.217 g/mol), is dissolved in 50.0 g of acetone (MW = 58.09 g/mol).

Calculate the vapor pressure of acetone above the resulting solution.

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

solution
Solution

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

problem1
Problem

The vapor pressure of pure liquid CS2 is 0.3914 atm at 20°C. When 40.0 g of rhombic sulfur (a naturally occurring form of sulfur) is dissolved in 1.00 kg of CS2, the vapor pressure falls to 0.3868 atm.

Determine the molecular formula of rhombic sulfur.

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide15
Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร
boiling pt elevation freezing pt depression
Boiling Pt. Elevation Freezing Pt. Depression

ΔTb = T boiling, solution – T boiling, pure solvent = Kb m

ΔTf = T freezing, solution – T freezing, pure solvent = - Kf m

m = molality of the solution

Kb = boiling constant

Kf = cryoscopic constant

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide17

Freezing is a dynamic equilibrium between melting and freezing.

P

T = K

P = atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide18

Freezing is a dynamic equilibrium between melting and freezing.

P

T = 0 oC

P = 1atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

boiling point elevation
Boiling Point Elevation

At equilibrium, the rate of evaporation (liquid to gas) equals the rate of condensation (gas to liquid)

P

T = 100 oC

P = 1 atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

boiling point elevation1
Boiling Point Elevation

P

T = 100 oC

P = 1 atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

boiling point elevation2
Boiling Point Elevation

P

T = K

P = atm

Surface of liquid

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide22

0

DTb = Tb – T b

0

T b is the boiling point of

the pure solvent

0

Tb > T b

DTb = Kbm

Boiling-Point Elevation

T b is the boiling point of

the solution

DTb > 0

m is the molality of the solution

Kb is the molal boiling-point

elevation constant (0C/m)

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide23

0

DTf = T f – Tf

0

T f is the freezing point of

the pure solvent

0

T f > Tf

DTf = Kfm

Freezing-Point Depression

T f is the freezing point of

the solution

DTf > 0

m is the molality of the solution

Kf is the molal freezing-point

depression constant (0C/m)

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide24
Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร
boiling pt elevation freezing pt depression1
Boiling Pt. Elevation Freezing Pt. Depression

ΔTb = T boiling, solution – T boiling, pure solvent = Kb m

ΔTf = T freezing, solution – T freezing, pure solvent = - Kf m

m = molality of the solution

Kb = boiling constant

Kf = cryoscopic constant

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide26

0

DTf = T f – Tf

moles of solute

m=

mass of solvent (kg)

=

3.202 kg solvent

1 mol

62.01 g

478 g x

0

Tf = T f – DTf

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

DTf = Kfm

Kf water = 1.86 0C/m

= 2.41 m

DTf = Kfm

= 1.86 0C/m x 2.41 m = 4.48 0C

= 0.00 0C – 4.48 0C = -4.48 0C

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide27

Vapor-Pressure Lowering

Boiling-Point Elevation

DTb = Kbm

0

P1 = X1 P 1

Freezing-Point Depression

DTf = Kfm

p = MRT

Osmotic Pressure (p)

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.

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide28
How many grams of sucrose (C12H22O11) are needed to lower the freezing point of 100 g of water by 3° C?

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide29
ΔTf = - Kf m

We want to decrease the freezing point by 3°C

-3° C = -(1.86 °C/molal) m

m=1.61 molal = 1.61 moles solute/kg solvent

NOTE: Kf is the WATER cryoscopic constant

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide30
1.61 moles solute = x moles solute

1 kg solvent 0.100 kg water

0.161 moles sucrose x 342 g sucrose = 55.1 g sucros 1 mole sucrose

342 g/mol = 11*12.01 g/mol + 22x1.008 g/mol + 11x16 g/mol

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide31
How many grams of NaCl are needed to lower the freezing point of 100 g of water by 3 °C?

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

the answer
The Answer

ΔTf = = - Kf m

We want to decrease the freezing point by 3°C

-3° C = -(1.86 °C/molal) m

m=1.61 molal = 1.61 moles solute/kg solvent

NOTE: Kf is the WATER cryoscopic constant

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide33
1.61 moles solute * 0.100 kg water = 0.161 moles solute

1 kg solvent

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

it s all about the of particles
It’s all about the of particles

1.61 moles solute * 0.100 kg water = 0.161 moles solute

1 kg solvent

BUT NaCl is an electrolyte:

NaCl Na+ + Cl-

You get 2 moles of solute per mole NaCl

0.161 moles solute * 1 mol NaCl * 58.45 g NaCl = 4.7 g NaCl 2 mol solute 1 mole NaCl

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide35

actual number of particles in soln after dissociation

van’t Hoff factor (i) =

number of formula units initially dissolved in soln

Colligative Properties of Electrolyte Solutions

0.1 m NaCl solution

0.1 m Na+ ions & 0.1 m Cl- ions

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

0.1 m NaCl solution

0.2 m ions in solution

i should be

1

nonelectrolytes

2

NaCl

3

CaCl2

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide36

Boiling-Point Elevation

DTb = iKbm

Freezing-Point Depression

DTf = i Kfm

p = iMRT

Osmotic Pressure (p)

Colligative Properties of Electrolyte Solutions

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

freezing point depression
At what temperature will a 5.4 molal solution of NaCl freeze?

Solution

∆TFP = Kf • m • i

∆TFP = (1.86 oC/molal) • 5.4 m • 2

∆TFP = 20.1oC

FP = 0 – 20.1 = -20.1oC

Freezing Point Depression

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

osmotic pressure
Osmotic Pressure

Osmotic pressure is the “funky” colligative property, but it is very important biologically

Osmotic pressure is the pressure required to overcome the natural pressure exerted by a solution by virtue of having a concentration.

Osmotic pressure looks just like the ideal gas law:

ΠV = nRT

where Π is osmotic pressure

Π= (n/V) RT = M RT

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide39

What Happens?

1 M NaCl

4 M NaCl

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide40

The water moves from the 1 M side

to the 4 M side. Why?

1 M NaCl

4 M NaCl

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide41
OSMOTIC PRESSURE

Π= M RT

On the 1 M side, the pressure is:

Π= M RT = 1 M (0.0821 Latm/mol K)(298 K)

Π=24.4 atm

On the 4 M side, the pressure is:

Π= M RT = 4 M (0.0821 Latm/mol K) (298 K)

Π= 97.9 atm

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide42

Both solutions “push” on the membrane

The bigger push wins!

1 M NaCl

4 M NaCl

24.4 atm

97.9 m

73.5 atm

Note the direction of the arrows. Osmotic pressure is pushing AGAINST the solution.

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide43

Osmotic Pressure (p)

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 (p) is the pressure required to stop osmosis.

more

concentrated

dilute

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide44

Osmotic Pressure (p) in air

High

P

Low

P

p = MRT

M is the molarity of the solution

R is the gas constant

T is the temperature (in K)

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide45

A cell in an:

isotonic

solution

hypotonic

solution

hypertonic

solution

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide46

Chemistry In Action:

Desalination

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร

slide47
แหล่งอ้างอิง
  • Martin S. Silberberg, Chemistry: The Molecular Nature of Matter and Change, McGraw-Hill Higher Education, 2004
  • Raymond Chang, Chemistry, Williams College, McGraw-Hill Higher Education, 2002

Colligative properties ผู้สอน: อ.ศราวุทธ แสงอุไร