Chap13_Part4

Chap13_Part4 - 1)
Mass
Percentage
(wt%)


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Unformatted text preview: 1)
Mass
Percentage
(wt%)
 For
dilute
solu,ons,
some,mes
the
mass
%
is
referred
to

 as
“parts
per
million”:
 mass of component ppm = x 10 6 total mass of solution For
very
dilute
solu,ons,
some,mes
the
mass
%
is
referred
to

 as
“parts
per
billion”:
 mass of component ppb = x 10 9 total mass of solution For
dilute
and
very
dilute
solu,ons,
the
total
mass
 of
the
solu,ons
is
≈
the
total
mass
of
the
solvent.
 2
 2)
Mol
Frac5on
 mols of component mol fraction of component = X = total mols in solution 3)
Molarity
 mols of solute molarity = M = liters of solution 4)
Molality
 mols of solute molality = m = kg solvent When
the
solvent
is
water,
1
L
≈
1
kg.

Thus,
when
the
solu,on
is
 dilute,
the
denominators
of
3)
and
4)
are
equivalent.
 3
 Calcula5ng
Molarity
and
Molality
 
If
we
know
the
density
 of
the
solu,on,
we
 can
calculate
the
 molality
from
the
 molarity
and
vice
 versa.
 4
 A
solu,on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac,on;

 (c)
molality;
and
(d)
molarity.


 (a)
 5
 A
solu,on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
g/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac,on;

 (c)
molality;
and
(d)
molarity.


 (a)
 mass solute 80.5 x100 = = 27.7% mass solution 80.5 + 210 (b)
 6
 A
solu,on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac,on;

 (c)
molality;
and
(d)
molarity.


 (a)
 (b)
 mass solute 80.5 x100 = = 27.7% mass solution 80.5 + 210 molecular weight solute = 6(12) + 8(1) + 6(16) = 176 mol solute = gm mol 80.5 gm = 0.457 mol 176 gm / mol molecular weight solvent = 18 gm mol 210 gm = 11.7 mol 18 gm / mol 0.457 mol fraction = = 0.038 11.7 + 0.457 mol solvent = 7
 A
solu,on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac,on;

 (c)
molality;
and
(d)
molarity.


 (c)
 8
 A
solu,on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac,on;

 (c)
molality;
and
(d)
molarity.


 (c)
 mol solute = 0.457 mol mass solvent = 210 gm = 0.210 kg 0.457 mol mol molality = = 2.18 0.210 kg kg (d)
 9
 A
solu,on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac,on;

 (c)
molality;
and
(d)
molarity.


 (c)
 mol solute = 0.457 mol mass solvent = 210 gm = 0.210 kg 0.457 mol mol molality = = 2.18 0.210 kg kg (d)
 mol solute = 0.457 mol (210 + 80.5) gm volume solution = = 238 mL = 0.238 L 1.22 gm / mL 0.457 mol mol molarity = = 1.92 0.238 L L 10
 Colliga5ve
Proper5es
 Changes
in
colliga,ve
proper,es
depend
only
on
the
number
of

 solute
par,cles
present,
not
on
the
iden)ty
of
the
solute
par,cles.
 These
proper,es
are
called
“colliga,ve”:
 
 
1)
vapor
pressure
lowering
 
 
2)
boiling
point
eleva,on
 
 
3)
freezing
point
depression
 
 
4)
osmo,c
pressure
 Lowering
the
Vapor
Pressure
 A
liquid
in
a
closed
container
will
come
into
 equilibrium
with
its
vapor.
 At
that
point,
the
gas
pressure
is
called
the
 Vapor
pressure.
 Substances
that
have
no
measurable
vapor

 pressures
are
nonvola,le;
substances
that

 do
are
vola,le.
 The
vapor
pressure
of
a
vola,le
solvent
containing
 a
nonvola,le
solute
is
lower
than
the
vapor

 pressure
of
the
pure
solvent.
 The
amount
by
which
the
vapor
pressure
is
 
reduced
depends
on
the
concentra,on
of
solute.
 The
solute
pulls
solvent
into
the
solu,on.
 Raoult’s
Law:

Vapor
Pressure
Decrease
with
Solute
 “The
vapor
pressure
of
a
gas
above
a
solu,on
equals
the
product
of
the
mol

 frac,on
of
solvent
in
the
solu,on
and
the
vapor
pressure
of
the
pure
solvent”.
 PA = X A PA 0 PA
=
vapor
pressure
of

 gas
above
solu,on

 XA
=
mol
frac,on
of

 solvent
in
solu,on
 PA0
=
vapor
pressure
 of
pure
solvent
 XA
usually
does
not
differ
too
much
from
one
and
the
effects
are
usually
small.
 Francois‐Marie
Raoult
(1830‐1901)
 ‐ 
French
Chemist
 ‐ 
Raoults
law
 ‐ 
Work
on
freezing
point
depression
 ‐ Lowering
the
vapor
pressure
 ‐ PhD
work
on
electrochemistry
 The
vapor
pressure
of
pure
water
at
70oC
is
0.31
atm.

Calculate
the
vapor

 pressure
of
125
gm
water
containing
35.0
gm
glycerin
(C3H8O3)
in
water

 at
this
temperature.

 The
vapor
pressure
of
pure
water
at
70oC
is
0.31
atm.

Calculate
the
vapor

 pressure
of
125
gm
water
containing
35.0
gm
glycerin
(C3H8O3)
in
water

 at
this
temperature.

 molecular weight of glycerin = 3(12) + 8(1) + 3(16) = 92 35.0 gm mol solute = = 0.38 mol 92 gm / mol 125 gm mol solvent = = 6.94 mol 18 gm / mol 6.94 mol fraction solvent = = 0.95 6.94 + 0.38 0 PA = X A PA = (0.95)(0.31 atm) = 0.29 atm gm mol Although
the
effects
are
small,
the
lowering
of
vapor
pressure
can
be
used
to
determine
solute
 concentra,ons.

At
the
top,
the
solvent
is
1
kg
≈
1
L
H2O
and
the
height
difference
is
17.54
mm.


 At
the
bobom,
a
certain
amount
of
the
sugar
mannitol
(182.2
gm/mol)
has
been
added
and

 the
height
difference
is
17.45
mm
(requiring
something
other
than
a
ruler
to
measure).


 Assume
that
the
volume
does
not
change.

Find
the
concentra,on
of
mannitol
in
units
of
gm/L.
 barometer
=
closed
top
 Although
the
effects
are
small,
the
lowering
of
vapor
pressure
can
be
used
to
determine
solute
 concentra,ons.

At
the
top,
the
solvent
is
1
kg
≈
1
L
H2O
and
the
height
difference
is
17.54
mm.


 At
the
bobom,
a
certain
amount
of
the
sugar
mannitol
(182.2
gm/mol)
has
been
added
and

 the
height
difference
is
17.45
mm
(requiring
something
other
than
a
ruler
to
measure).


 Assume
that
the
volume
does
not
change.

Find
the
concentra,on
of
mannitol
in
units
of
gm/L.
 XA = PA PA 0 = 17.45 mm = 0.9949 17.54 mm 1000 gm = 55.56 mol 18 gm / mol n solvent XA = ⇒ X A (n solvent + n solute ) = n solvent n solvent + n solute n solvent = ⇒ barometer
=
closed
top
 ⇒ n solute = (1 − 0.9949)(55.56 mol ) = 0.2848 mol 0.9949 182.2 gm = (0.2848 mol )( ) = 51.89 gm mol gm ⇒ c solute = 52 L n solute = m solute X A n solute = (1 − X A )n solvent (1 − X A )n solvent XA ‐ 
Solu,ons
that
obey
Raoult’s
law
:
Ideal
Solu,ons
 ‐ 
Devia,ons
from
the
Ideal
behavior
of
the
solu,ons:
Based
on
 

intermolecular
forces.

 ‐ Strong
Solvent/Solvent

or
Solute/Solute
interac,ons:
Vapor
 
pressure
is
more
than
predicted
by
Raoult’s
Law.
 (Example:
Actual
P=
1
atm,
and
calculated
P
according
to
Raoult’s
law
ager
the

 Addi,on
of
solute
is
0.95
atm‐
you
will
find
PA>0.95
but
<
1)
 ‐ Strong
Solute/Solvent
interac,ons:
Vapor
pressure
is
less
than
 
predicted
by
Raoult’s
law.
 (Example:
Actual
P=
1
atm,
and
calculated
P
according
to
Raoult’s
law
ager
the

 Addi,on
of
solute
is
0.95
atm‐
you
will
find
find
PA<0.95)
 IMPORTANT:
EXAM
 ‐ 
Review
session
TODAY
7
pm
onwards
 ‐ 
You
need
Blue/
Black
pen
for
the
exam:

Pencils
are
not
allowed
 ‐ 
You
need
CALCULATORS:
Using
Computers/
Cell
phones
is
prohibited
 








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sure
they
are
working
well)
 ‐ 
What
is
the
best
way
to
do
well
in
the
exam?
 ‐ 
Read
the
problem
once/
twice
 ‐ 
Do
not
spend
on
any
problem
more
than
3‐4
minutes
 ‐ 
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sure
to
write
clearly

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