larutan 62-65
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7/26/2019 larutan 62-65
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Campuran 60 g air + 40 g fenol pada 300C
Lapisan air (A) = H2O = 92 %
enol = ! %
Lapisan fenol (") = H2O = 30 %
enol = #0 %
$ian&a'an era lapisan air* A = ,
era lapisan fenol* " = ,
Auran perandingan
W lap . H 2O
W lap . fenol =CB
CA
berat lapisan air
berat lapisan fenol
=W A
W B=¿
CB
CA
W A
Wt −W A=
CB
CA
A + " = oal
Wt −W A
W A =
CA
CB
Wt
W A - =
CA
CB
W A =
Wt
CACB +1 . prolema sandar
W B = -00 /
W A
W A =
100
40−8
70−40+1 g
W B = -00 / 4!*4 g
W A =
100
32
30+1 =
3000
62
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W A = 4!*4
W B = -*6
1OAL 9
1OAL 6 60
Laruan -0 g a dalam 300 g CCl4 (ideal)
5ada 50 = - am . ii' didi7 * = 0 + 8d
5ada 5 = 0*99# am . ii' didi7 = 0 ( ii' diidi7
normal CCl4)
$ian&a'an 2 = ,
8 d = :d m
:d = ; 0 2 - = ( li7a 7al 3!)
-00 0 8H<
d ln P
dT =∆Hv
RT 2
( pers Cla&peron Clausius )
m = Mol zat kg pelarut
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m =
W 2
M 2
1000
W 1 . 2 =
W 2
W 1 1000
m ( m ida' di'ea7ui )
8d = :d m . m =∆Td
d
:d = R T 0
2 M 1
1000 ∆ Hv
d ln P
dT =∆Hv
RT 2 . ln
P0
P =∆Hv
R T −T
0
T .T 0 >
∆ Hv
R !∆Td
T 02
5engandaian 8H< ? f ()
=∆ Hv
R d . M
T 02
=∆Hv
R @ m
T 02 @
RT 02 M 1
1000∆Hv
ln P
0
P =m. M 1
1000 . m =1000
M 1 ln P
0
P
1usiusi 2 = "2
W 1
1000
1000
M 1. ln
P0
P . 2 =
W 2
W 1
M 1
ln P
0
P . pro sandar
=
10
300 .
153,81
ln 1
0,97
2 = -#0*9 g ol -
1OAL # 6-
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Laruan 3 g a dalam -00 g CCl4
8d = 0*600
:d = *03
: = 3-*!
$ian&a'an (a) 8 = ,
() 8 P
P0 = , penurunan e' uap 'eluar
() # pada 29!0 : = , (am)
(d) 2 = ,
8d = :d m
8 = : m
85 = 2 50
# = ;
m =mol zat
kg pelarut
8 8 = : m
m 8d = :d m . m =∆Td
d . prolema sandar
∆ P
P0 85 = 2 5
0 .∆ P
P0 = 2 =
n2
n1+n2 >
n2
n1 >
"2
M 2
W 1
M 1
=
W 2
M 2 @
M 1
"1
m =
W 2
M 2 @1000
"1 =
∆ Td
d .W
2
M 2 =
W 1
1000 @∆Td
d
susiusi
∆ P
P0 =
W 1
1000 @∆ Td
d @ M 1
"1 .
∆ P
P0 =
∆ T d . M
1000 . d .
prol s
# # = ; > m ; ( 'arena lar Bner) . # =∆Td
d @ ; . pro s
2 m =
W 2
M 2 @1000
"1 . 2 =
W 2
m @1000
"1 =
W 2
∆Td
d
@1000
"1 .
pro s
8 = :
∆Td
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(a) 8 = 3-*! 0,60
5,03
()∆ P
P0 =
(0,60 )(153,81)
(1000 )(5,03)
() # =
(0,60 ) (0.0821 )(298)5,03
(d) 2 =
3
0,60
5,03
1000
100
(a) 8 = 3*#90
()
∆ P
P0 = 0*0-!
() # = 2*92 am
(d) 2 = 2-* g ol-
1OAL ! 62
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Laruan su'rosa dalam air " = 0*2000 C
5H20 = 23*#6 orr (20C)
: = -*!6 0C molal-
$ian&a'an 5lar pada 20C = , ( 5H20 )
8 = : m
8 = O "
85 = 2 50
5 85 = 2 50 . 50 / 5 = 2 5
0 . 5 = 50 ( - 2) (2 ida' di'e)
2 8" = : m . m =∆ Tb
b molal = ∆Tb
b ol per 1000
M H 20 ol H20
2 =
∆Tb
b
∆Tb
b +
1000
MH 2O
=
∆Tb
∆Tb+1000
MH 2O
1usiusi 5 = 50 -
∆Tb
∆Tb+1000
MH 2O . prolema sandar
5 = 23*#6 / - /
0,200
0,200+1000 !1,86
18,02
orr
5 = 23*#6 ( - / 0*00-93 )
= 23*#-0 orr