Colligative Properties of Solutions Colligative properties = physical properties of solutions that...
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Colligative Properties of Solutions Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.
Colligative Properties of Solutions Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind
Colligative Properties of Solutions Colligative properties =
physical properties of solutions that depend on the # of particles
dissolved, not the kind of particle.
Slide 3
Colligative Properties n Lowering vapor pressure n Raising
boiling point n Lowering freezing point n Generating an osmotic
pressure
Slide 4
Colligative Properties n Lowering vapor pressure n Raising
boiling point n Lowering freezing point n Generating an osmotic
pressure
Slide 5
Boiling Point Elevation n a solution that contains a
nonvolatile solute has a higher boiling pt than the pure solvent;
the boiling pt elevation is proportional to the # of moles of
solute dissolved in a given mass of solvent.
Slide 6
Boiling Point Elevation n T b = k b m where: T b = elevation of
boiling pt m = molality of solute k b = the molal boiling pt
elevation constant n k b values are constants; see table 15-4, p.
472 (honors text) n k b for water = 0.52 C/m
Slide 7
Ex: What is the normal boiling pt of a 2.50 m glucose, C 6 H 12
O 6, solution? n normal implies 1 atm of pressure n T b = k b m n T
b = (0.52 C/m)(2.50 m) n T b = 1.3 C n T b = 100.0 C + 1.3 C =
101.3 C
Slide 8
Ex: How many grams of glucose, C 6 H 12 O 6, would need to be
dissolved in 535.5 g of water to produce a solution that boils at
101.5C? n T b = k b m n 1.5 C= (0.52 C/m)(m) n m = 2.885
Slide 9
Freezing/Melting Point Depression n The freezing point of a
solution is always lower than that of the pure solvent.
Slide 10
Freezing/Melting Point Depression n T f = k f m where: T f =
lowering of freezing point m = molality of solute k f = the
freezing pt depression constant n k f for water = 1.86 C/m n k f
values are constants; see table 15-5, p. 474 (honors text)
Slide 11
Ex: Calculate the freezing pt of a 2.50 m glucose solution. n T
f = k f m n T f = (1.86 C/m)(2.50 m) n T f = 4.65 C n T f = 0.00 C
- 4.65 C = -4.65 C
Slide 12
Ex: When 15.0 g of ethyl alcohol, C 2 H 5 OH, is dissolved in
750 grams of formic acid, the freezing pt of the solution is 7.20C.
The freezing pt of pure formic acid is 8.40C. Determine K f for
formic acid. T f = k f m 1.20 C= (k f )( 0.4348 m) k f = 2.8
C/m
Slide 13
Ex: An antifreeze solution is prepared containing 50.0 cm 3 of
ethylene glycol, C 2 H 6 O 2, (d = 1.12 g/cm 3 ), in 50.0 g water.
Calculate the freezing point of this 50-50 mixture. Would this
antifreeze protect a car in Chicago on a day when the temperature
gets as low as 10 F? (-10 F = -23.3 C) T f = k f m T f = (1.86
C/m)(18.06 m) T f = 33.6 C T f = 0 C 33.6 C = -33.6 C YES!
Slide 14
Electrolytes and Colligative Properties Colligative properties
depend on the # of particles present in solution. Because ionic
solutes dissociate into ions, they have a greater effect on
freezing pt and boiling pt than molecular solids of the same molal
conc.
Slide 15
Electrolytes and Colligative Properties n For example, the
freezing pt of water is lowered by 1.86C with the addition of any
molecular solute at a concentration of 1 m. Such as C 6 H 12 O 6,
or any other covalent compound n However, a 1 m NaCl solution
contains 2 molal conc. of IONS. Thus, the freezing pt depression
for NaCl is 3.72Cdouble that of a molecular solute. NaCl Na + + Cl
- (2 particles)
Slide 16
Electrolytes - Boiling Point Elevation and Freezing Point
Depression The relationships are given by the following equations:
n T f = k f mn or T b = k b mn T f/b = f.p. depression/elevation of
b.p. m = molality of solute k f/b = b.p. elevation/f.p depression
constant n = # particles formed from the dissociation of each
formula unit of the solute
Slide 17
Ex: What is the freezing pt of: a) a 1.15 m sodium chloride
solution? n NaCl Na + + Cl - n=2 n T f = k f mn n T f = (1.86
C/m)(1.15 m)(2) n T f = 4.28 C n T f = 0.00 C - 4.28 C = -4.28
C
Slide 18
Ex: What is the freezing pt of: b) a 1.15 m calcium chloride
solution? n CaCl 2 Ca 2+ + 2Cl - n=3 n T f = k f mn n T f = (1.86
C/m)(1.15 m)(3) n T f = 6.42 C n T f = 0.00 C 6.42 C = -6.42 C
Slide 19
Ex: What is the freezing pt of: c) a 1.15 m calcium phosphate
solution? n Ca 3 (PO 4 ) 2 3Ca 2+ + 2PO 4 3- n n=5 n T f = k f mn n
T f = (1.86 C/m)(1.15 m)(5) n T f = 10.7 C n T f = 0.0 C 10.7 C =
-10.7 C
Slide 20
Determining Molecular Weights by Freezing Point Depression
Slide 21
n T f = 0.56C n T f = k f m n 0.56C = (5.12C/m)(m) n m = 0.1094
Ex: A 1.20 g sample of an unknown molecular compound is dissolved
in 50.0 g of benzene. The solution freezes at 4.92C. Determine the
molecular weight of the compound. The freezing pt of pure benzene
is 5.48C and the K f for benzene is 5.12C/m.
Slide 22
Ex: A 37.0 g sample of a new covalent compound was dissolved in
200.0 g of water. The resulting solution froze at 5.58C. What is
the molecular weight of the compound? n T f = 5.58C n T f = k f m n
5.58C = (1.86C/m)(m) n m = 3.00 m