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University of Illinois at ChicagoUICCHEM 112: General Chemistry I Lecture

Instructor: Dr. Chad LandrieLecture CRN: 18644

Time/Day: M,W,F; 2:00-2:50 pmMarch 15, 2010

General Chemistry 1Lecture 27

UICUniversity of Illinois at Chicago 2010, Dr. Chad L. LandrieCHEM 112: General Chemistry 1, Spring 2010 SlideLecture: March 15

i>Clicker Question

2

Determine the vapor pressure of an aqueous solution that is 5.00% (w/w) NaCl if the equilibrium vapor pressure of water is 27.0 torr.

PH2O = PH2Oo 2

5 % (w/w) = 5 g NaCl95 g H2O

5 g NaCl

1 mol NaCl58.44 g NaCl

= 0.08556 mol NaCl

95 g H2O 1 mol H2O18.016 g H2O

= 5.273 mol H2O

PH2O = (27.0 torr) (0.9686) = 26.2 torr

A. 25.7B. 26.6C. 26.2

D. 0.862E. none

2 =nH2Ontotal

=5.273

2(0.08556) + 5.273= 0.9686


i>Clicker Question

3

A 5.00% (w/w) aqueous solution of acetone (C3H6O) is prepared. The equilibrium vapor pressures of water and acetone are 24.0 and 350 torr, respectively. Determine the total vapor pressure of the solution.

95 g H2O 1 mol H2O18.016 g H2O

= 5.273 mol H2O

A. 374B. 0.0206C. 43.2D. 33.5E. none

5.00 g C3H6O 1 mol C2H6O58.078 g C2H6O

= 0.08609 mol C3H6O

2 =nH2Ontotal

=5.273

0.08609 + 5.273= 0.9839 mol C2H6O

PH2O = (24.0 torr) (0.9839) = 23.6136 torrPC3H6O = (350. torr) (1-0.9839) = 5.635 torrPtot = 23.6136 torr + 5.635 torr = 29.2486 torr


i>Clicker Question

4

A. 0.174B. 0.0202C. 0.980D. 0.212E. none

A 5.00% (w/w) aqueous solution of acetone (C3H6O) is prepared. The equilibrium vapor pressures of water and acetone are 24 and 350 torr, respectively. Determine the total vapor pressure of the solution. Then, determine the mole fraction of acetone in the vapor.

2 =nH2Ontotal

=5.273

0.08609 + 5.273= 0.9839 mol C2H6O

PH2O = (24.0 torr) (0.9839) = 23.6136 torrPC3H6O = (350. torr) (1-0.9839) = 5.635 torrPtot = 23.6136 torr + 5.635 torr = 29.2486 torr

95 g H2O 1 mol H2O18.016 g H2O

= 5.273 mol H2O5.00 g C3H6O 1 mol C2H6O

58.078 g C2H6O= 0.08609 mol C3H6O

C2H6O in vapor =PC3H6O

PC3H6O + PH2O

C2H6O in vapor =5.635

29.2486=


Vapor Depression (P)

5

Case 1: Nonvolatile, NonelectrolytesThe equilibrium vapor pressure for a liquid decreases as the concentration of solute increases.

Mathematically, the partial vapor pressure of the solvent (Psolvent) depends on its mole fraction (X) in the solution.

Xsolvent =nsolventntotal

=nsolvent

nsolvent + nsolute

Psolvent = Xsolvent Psolvento

P = partial pressure (mixtures)P = equilibrium vapor pressre (pure liquids)

mole fraction

RaoultsLaw

P = Psolvento Psolvent

P = Xsolute( ) Psolvento( )



6

Case 2: Nonvolatile, Strong ElectrolytesThe equilibrium vapor pressure for a liquid decreases as the concentration (total for all ions) of solute increases.

12 mol NaCl 4800 mol H2O 12 mol Na+12 mol Cl

+

Xsolute = nNa+ + nCl( )

nNa+ + nCl( ) + nH2O( )=

12 +12(12 +12)+ 4800

= 0.005

The total moles of solute is the sum of the moles of each dissociated ion.



7

Case 3: Volatile, NonelectrolytesThe total vapor pressure for a solution is the sum (Daltons Law) of the partial pressures for each component (Raoults Law).

vapo

r pr

essu

re (

torr

)

mole fraction (in the solution)

XAXB 0

1100.5

0.50.25 0.750.250.75

Ptotal = PA + PB

PA = XA PAo

PB = XB PBo

Ptotal

PA

PB

PA

PB

in soln =nA

nA + nB

XA in vapor =PA

PA +PB


Boiling Point

8

The boiling point of a liquid is the temperature at which its equilibrium vapor pressure (P) is equal to the atmospheric pressure (Patm). At this point there is rapid evaporation throughout the liquid, not just at the surface.

bp: Patm = PX

Therefore, anything that lowers P of a liquid (e.g., increased IMF), raises the boiling point since more energy (higher T) is required to obtain a P equal to Patm. Likewise anything the raises P of a liquid (e.g., decreased IMF), lowers the boiling point.


Boiling Point Elevation (Tb)

9

Tb = kbm

Tb = Tb (solution) Tb (solvent)

Tb = boiling point (C)kb = molal b.p. elevation constantm = molality ( mol solutekg solvent )

specific for each solvent; does not depend on the solute

must be molality for kb; not molarity


Freezing Point Depression (Tf)

10

specific for each solvent; does not depend on the solute

must be molality for kb; not molarity

Tf = kfm Tf = Tf (solution) Tf (solvent)

Tf = freezing point (C)kf = molal f.p. depression constantm = molality ( mol solutekg solvent )


Osmotic Pressure ()

11

more concentrated solution = higher in energy

osmosis: solvent will diffuse through a semipermeable membrane in the direction that reduces the concentration of the most concentrated solution

osmotic pressure () = pressure required to prevent osmosis

hypertonic = more concentrated solution

hypotonic = less concentrated solution


Osmotic Pressure ()

12

= nsoluteVsolution

RT

= MRT

= osmotic pressure (atm)nsolute = moles soluteVsolution = volume of solution (L)R = gas constant; 0.08206 LatmmolK( )T = temperatuer (K)


Determining Molar Mass

13

Tb = kbm





= nsoluteVsolution

RT

= MRT = osmotic pressure (atm)nsolute = moles soluteVsolution = volume of solution (L)R = gas constant; 0.08206 LatmmolK( )T = temperatuer (K)

Since all colligative properties relate number of particles (moles) to a change () in a measurable quantity, they can be used to determine molar mass (g/mol)

of unknown compounds if the mass used to prepare the solution is known.



14

Tb = kbm





= nsoluteVsolution

RT


measure Tknow kg solvent

know unknown mass (g)find mole solvent

calculate MM (g/mol)



15

Tb = kbm





= nsoluteVsolution

RT


measure Tknow kg solvent

know unknown mass (g)find mole solvent




16

Tb = kbm





= nsoluteVsolution

RT


measure know T, V, R & mass solute (g)

find concentration (M)calculate moles solute (mol)



i>Clicker Question

17

Which of the following solutions will have the highest osmotic pressure?

Colligative properties depend on the total

number of solutes, not the identity.


i>Clicker Question

18

431. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.325 atm at 25.0 C. Determine the molar mass of the protein.

=nproteinVsoln

RT

nprotein =VsolnRT

=(0.325 atm)(0.00500 L)

(0.08206 LatmmolK )(298.15 K) = 6.6418 105 mol

MM = m (g)n (mol)

=0.431 g

6.6418 105= 6490 g/mol

A. 6.49 x 103

B. 2.46 x 102

C. 343

D. 686

E. none

University of Illinois at ChicagoUICCHEM 112: General Chemistry I Lecture

Sections: 17.1 & 17.2

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