Review Session-5

Review Session-5 - REVIEW
SESSION‐
5
 Proper1es
of
Solu1ons


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Unformatted text preview: REVIEW
SESSION‐
5
 Proper1es
of
Solu1ons
 13.60)
(a)
What
is
ideal
solu4on?
(b)
The
vapor
pressure
of
pure
 water
at
60oC
is
149
torr.
The
vapor
pressure
of
water
over
a
solu4on
 containing
equal
number
of
moles
of
water
and
ethylene
glycol
is
67
 torr.
Is
the
solu4on
is
ideal
according
to
Raoults
law?
Explain?
 13.60)
(a)
What
is
ideal
solu4on?
(b)
The
vapor
pressure
of
pure
 water
at
60oC
is
149
torr.
The
vapor
pressure
of
water
over
a
solu4on
 containing
equal
number
of
moles
of
water
and
ethylene
glycol
is
67
 torr.
Is
the
solu4on
is
ideal
according
to
Raoults
law?
Explain?
 PA = X A PA 0 PA
=
vapor
pressure
of

 gas
above
solu4on

 XA
=
mol
frac4on
of

 solvent
in
solu4on
 PA0
=
vapor
pressure
 of
pure
solvent
 13.60)
(a)
What
is
ideal
solu4on?
(b)
The
vapor
pressure
of
pure
 water
at
60oC
is
149
torr.
The
vapor
pressure
of
water
over
a
solu4on
 containing
equal
number
of
moles
of
water
and
ethylene
glycol
is
67
 torr.
Is
the
solu4on
is
ideal
according
to
Raoults
law?
Explain?
 (a)  Ideal
solu4on
is
that
obeys
Raoults
law
 PA = X A PA 0 (b)  
Number
of
moles
of
water
=
number
of
moles
of
Ethylene
glycol
 XA
=
0.5
,

PA
=
(0.5)
(149)
=
74.5mm
of
Hg
 This
solu4on
has
a
vapor
pressure
that
is
higher
than
the
value
 obtained
experimentally
(67
torr).
Theore4cal
Pressure
(Raoults
 law)
is
more
than
observed
Pressure.
This
is
called
a
nega4ve
 devia4on.
 13.62)
(a)
Calculate
the
VP
of
water
above
a
solu4on
prepared
by
 dissolving
32.5g
of
glycerin(C3H8O3)
in
125g
of
water
at
343K.
(The
VP
 of
water
at
343K
=
233.7
torr)
(b)
Calculate
the
mass
of
ethylene
 glycol
(C2H6O2)
that
must
be
added
to
1kg
of
ethanol
(C2H5OH)
to
 reduce
the
VP
by
10
torr?
(The
VP
of
ethanol
at
35oC
is
100
torr)
 13.62)
(a)
Calculate
the
VP
of
water
above
a
solu4on
prepared
by
 dissolving
32.5g
of
glycerin(C3H8O3)
in
125g
of
water
at
343K.
(The
VP
 of
water
at
343K
=
233.7
torr)
(b)
Calculate
the
mass
of
ethylene
 glycol
(C2H6O2)
that
must
be
added
to
1kg
of
ethanol
(C2H5OH)
to
 reduce
the
VP
by
10
torr?
(The
VP
of
ethanol
at
35oC
is
100
torr)
 (a)  Number
of
moles
of
Glycerin
=
32.5/92.1
=
0.3529
mol
 





Number
of
moles
of
water
=
125/18
=
6.94

mol
 
Mol
frac4on
of
solvent
X
water=
6.94/(6.94+0.353)
=
0.952
 PH2O
=
0.952X233.7
=
222
torr


 13.62)
(a)
Calculate
the
VP
of
water
above
a
solu4on
prepared
by
 dissolving
32.5g
of
glycerin(C3H8O3)
in
125g
of
water
at
343K.
(The
VP
 of
water
at
343K
=
233.7
torr)
(b)
Calculate
the
mass
of
ethylene
 glycol
(C2H6O2)
that
must
be
added
to
1kg
of
ethanol
(C2H5OH)
to
 reduce
the
VP
by
10
torr?
(The
VP
of
ethanol
at
35oC
is
100
torr)
 (a)  Number
of
moles
of
Glycerin
=
32.5/92.1
=
0.3529
mol
 





Number
of
moles
of
water
=
125/18
=
6.94

mol
 
Mol
frac4on
of
solvent
X
water=
6.94/(6.94+0.353)
=
0.952
 PH2O
=
0.952X233.7
=
222
torr


 (b)
X
glycerin
=
10/100
=0.1
or
X
ethanol
=90/100
=0.9

 moles
of
ethanol
1000g/
46.07=
21.7
and

 “Y”
moles
of
Ethylene
glycol
   0.1
=
Y/Y+21.7

Y=
2.41
mol
   Mol.Wt
of
glycerin
=
62.07g/mol

62.07X2.41
=
150g
of
 ethylene
glycol
 13.66)
Arrange
the
following
aqueous
solu4ons,
each
10%
by
mass
in
 solute,
in
order
of
increasing
Boiling
Point?
Glucose
(C6H12O6),
 Sucrose
(C12H22O11),
NaNO3.

 13.66)
Arrange
the
following
aqueous
solu4ons,
each
10%
by
mass
in
 solute,
in
order
of
increasing
Boiling
Point?
Glucose
(C6H12O6),
 Sucrose
(C12H22O11),
NaNO3.

 10%

10g
in
100g
of
solu4on
 Number
of
moles:
Wt/
Mwt
 Higher
the
Mwt,
Lower
the
number
of
moles
&
Molecules.
 Sucrose
has
highest
MWt,
least
number
of
Molecules
 NaNO3:
Electrolyte:
can
give
2
ions:
More
molecules
 The
boiling
points
of
the
solu4ons
 
Sucrose
<
Glucose
<
NaNO3


 NaNO3
will
have
high
Boiling
Point

 13.71)
How
many
grams
of
ethylene
glycol
(C2H6O2)
must
be
added
to
 1kg
of
water
to
produce
a
solu4on
that
freezes
at
‐5oC
 13.71)
How
many
grams
of
ethylene
glycol
(C2H6O2)
must
be
added
to
 1kg
of
water
to
produce
a
solu4on
that
freezes
at
‐5oC
 ΔTf
=
5oC
=
Kf
(m)
 (m)
=
5oC/1.86
oC/m

=2.69
molals
of
C2H6O2
 m
=
mol
of
C2H6O2
/
kg
water
 Mol
of
C2H6O2=
m
x
1kg
water
=
2.69X1
=2.69moles
of
C2H6O2
 2.69
X
62.07
=
167g
of
C2H6O2
 13.72)
What
is
the
freezing
point
of
an
aqueous
solu4on
that
boils
at
 105oC?
 13.72)
What
is
the
freezing
point
of
an
aqueous
solu4on
that
bolis
at
 105oC?
 Kb
=0.51
and
Kf
=1.86
 Bp=105

ΔT=5oC
 ΔT
=
Kb
m

m=
ΔT/Kb
=
5/0.51
=
9.8m
 ΔTf
=
Kf
m
1.86
X9.8
=
18oC
 Freezing
point
=
0‐18
=
‐18oC


 13.74)
Sea
water
contains
3.4g
of
salts
for
every
liter
of
solu4on.
 Assuming
that
the
solute
consists
en4rely
NaCl,
calculate
the
osma4c
 pressure
of
sea
water
at
20oC
 13.74)
Sea
water
contains
3.4g
of
salts
for
every
liter
of
solu4on.
 Assuming
that
the
solute
consists
en4rely
NaCl,
calculate
the
osma4c
 pressure
of
sea
water
at
20oC
 Π =MRT,
T
=
293K
 M=
(3.4/58.5X1L)
X2
=
0.116M
 P
=
MRT
=
0.116
X
0.0821X
293K
=
2.8
atm
 13.78)
A
dilute
aqueous
solu4on
of
an
organic
compound
is
formed
 by
dissolving
2.35g
of
the
compound
in
water
to
form
0.35L
solu4on.
 The
resul4ng
solu4on
has
an
osma4c
pressure
of
0.605atm
at
 25oC
.Assuming
the
organic
compound
is
non
electrolyte,
what
is
the
 molar
mass?
 13.78)
A
dilute
aqueous
solu4on
of
an
organic
compound
is
formed
 by
dissolving
2.35g
of
the
compound
in
water
to
form
0.35L
solu4on.
 The
resul4ng
solu4on
has
an
osma4c
pressure
of
0.605atm
at
 25oC
.Assuming
the
organic
compound
is
non
electrolyte,
what
is
the
 molar
mass?
 Π =MRT,
T
=
298K
 M=
(0.605/0.0821X
298)=
0.0247M
 Molarity
=
wt/(Molwt
X
Vol
in
Liters)
 Mol.
Wt
=
Wt/(MolarityX
Vol
in
Liters)
 Mol
wt
=
2.35/(0.0247X0.35)
=
272g/mol

 13.78)
Fish
needs
at
least
4ppm
dissolved
oxygen
for
survival.
(a)
What
 is
this
concentra4on
in
mol/L
(b)
What
par4al
pressure
of
O2
above
the
 water
is
needed
to
obtain
this
concentra4on
at
10oC?
(Henry’s
law
 constant
for
O2
at
this
temperature
is
1.71X10‐3mol/L‐atm
 13.78)
Fish
needs
at
least
4ppm
dissolved
oxygen
for
survival.
(a)
What
 is
this
concentra4on
in
mol/L
(b)
What
par4al
pressure
of
O2
above
the
 water
is
needed
to
obtain
this
concentra4on
at
10oC?
(Henry’s
law
 constant
for
O2
at
this
temperature
is
1.71X10‐3mol/L‐atm
 (a)  4ppm
=

4mg
of
O2/1kg
of
solvent

 

















=
4X10‐3g/
1L
solvent
 
 





=
4X10‐3g/
(1L
solventX
32g
of
oxygen)
 
 





=
1.25X
10‐4M

 13.78)
Fish
needs
at
least
4ppm
dissolved
oxygen
for
survival.
(a)
What
 is
this
concentra4on
in
mol/L
(b)
What
par4al
pressure
of
O2
above
the
 water
is
needed
to
obtain
this
concentra4on
at
10oC?
(Henry’s
law
 constant
for
O2
at
this
temperature
is
1.71X10‐3mol/L‐atm
 (a)  4ppm
=

4mg
of
O2/1kg
of
solvent

 

















=
4X10‐3g/
1L
solvent
 
 





=
4X10‐3g/
(1L
solventX
32g
of
oxygen)
 
 





=
1.25X
10‐4M

 (b)  SO2=
kPO2
 
PO2=
SO2/k
=
1.25X10‐4
mol
/
1.71X10‐3
mol/L‐atm
 
PO2
=
0.0731
atm
=
60mm
of
Hg

 REVIEW
SESSION
:
SATURDAY:

 CHAPMAN
211
at
10.30
am
to
Noon

 ...
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This note was uploaded on 09/13/2011 for the course CHEM 102 taught by Professor Todd during the Spring '08 term at UNC.

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