lampiran a b fiizzx

8/19/2019 Lampiran a b Fiizzx

1/17

LAMPIRAN ADATA PERCOBAAN

A.1 Kalibrasi PelarutPelarut yang digunakan : Aquadest (H 2O)

Volume pelarut awal (V o) = 8 ml

Titik didih kalibrasi = ! o"

#ensitas pelarut ($ pelarut ) = %& '!2! g ml

Tekanan Pelarut murni (P o) = 8&8!8' kPa (*eankoplis& + !)

A.2 Penentuan Kenaikan Titik Didih den an Met!de Landsberger A.2.1 "a#$el % Malt!sa &C 12' 22O 11(Tabel ,-+ Hasil Per.obaan untuk /arutan 0altosa (" +2H22O++)

1u

n (g) V + (ml) V 2 (ml) Td ( %") 3Td ( %")

4 %&5 8&6 +! 5 +44 %&6 8&6 +! 5 +444 %&' 8&6 +5&6 6 24V %&8 8&6 +5&6 6 2

A.2.2 "a#$el % "ukr!sa &C 12 ' 22O 11(Tabel ,-2 Hasil Per.obaan untuk /arutan 7ukrosa (" +2H22O++)

1u

n (g) V + (ml) V 2 (ml) Td ( %") 3Td ( %")

4 %&5 8&6 ++&6 6 244 %&6 8&6 +2&6 ' !444 %&' 8&6 +2&6 ' !4V %&8 8&6 +! 5

A.2.) "a#$el % Tri*Natriu# P!s+at D!de,ahidrat &Na ) PO - .12' 2O( - NaO'(Tabel ,-! Hasil Per.obaan untuk /arutan Tri 9atrium Pos at #ode.ahidrat

(9a ! PO 5-+2H2O) 5 9aOH)

1un (g) V + (ml) V 2 (ml) Td ( %") 3Td ( %")

4 %&6 8 +2 ' !44 %&' 8 +2&6 5

444 %& 8 +! 8 6

Universitas Sumatera Utara /, +Kenaikan Titik Didih

Andri Rizki


2/17

/, 2

4V %& 8 +! 8 6

keterangan :

= massa ;at terlarut (g)V+ =


3/17

/, !

4nterpolasi untuk menentukan $ pada T = ! o":

+2

+

+2

+

TTTT

−

−=

−

−

ρ ρ

ρ ρ

=−

%+%%%!

%& '6!%& 685%& '6! ρ

=+%!

%&%%'%& '6! ρ

%&%%2% = +%$ > &'6!

+%$ = &'!26

$ = %& '!26 g ml

Pada T + = +%%o"& P+ = +%+&!26 kPa (*eankoplis& + !)

Pada T 2 = %o"& P2 = %&+% kPa (*eankoplis& + !)

4nterpolasi untuk menentukan P pada T = o":

+2

+

+2

+

TTTT

−

−=

−

− P P

P P

=−%+%%%!

)%&+%+%+&!26)%&+% P

=+%!

!+&226)%&+% P

!&' 6 = +%P > %+

+%P = 5&' 6

P = &5' 6 kP

Volume pelarut awal (V %) = 8 m/

Titik didih kalibrasi (T k ) = ! o"

#ensitas pelarut ($ pelarut ) = %& '!2! g m/

Tekanan uap pelarut (P? pelarut ) = 8&8!8' kPa (*eankoplis& + !)

0assa pelarut (P +) = ($ pelarut ) @ V%

= %& '!26 g m/ @ 8 m/

Universitas Sumatera UtaraKenaikan Titik Didih

Andri Rizki


4/17

/, 5

= & %' g

B.2 Perhitun an Penurunan Tekanan a$ Larutan

B.2.1 Penentuan 0raksi M!l at Terlarut

1. "a#$el % Malt!sa &C 12 ' 22O 11(

Run I

A0 " +2H22O++ = !52&!% g mol (7.ien.elab& 2%+!a)

A0 H 2O = +8&%2 g mol (7.ien.elab& 2%+!d)

OHA0

gramn

2+ =

'+2'2 OH"A0

gramn =

g(mol+8&%2g)&)%'

n+ = g(mol !%&!52g%&5

n 2 =

mol52)'&%n+ = mol%%+2&%n 2 =

+2

22 nn

nB

+=

mol%&52 'mol%%+2&%mol%&%%+2B 2 +

=

%%28&%B 2 =

#engan rumus tersebut diperoleh:

Tabel A-+ Hasil Penentuan Craksi 0ol 0altosa &" +2H22O++)

1un (g) P + n+ n2 B 24 %&5 & %' %&52 ' %&%%+2 %&%%28

44 %&6 %&%%+6 %&%%!6


Andri Rizki


5/17

/, 6

444 %&' %&%%+8 %&%%524V %&8 %&%%2! %&%%65

2. "a#$el % "ukr!sa &C 12 ' 22O 11( Run I

A0 " +2H22O+2 = !52&!% g mol (7.ien.elab& 2%+!b)


OHA0

gramn

2+ =

2

2 Aa"lA0

gramn =

g(mol+8&%2

g)&)%'n

+ =

g(mol !52&!%

g%&5n

2 =

mol52)'&%n+ = mol%%+2&%n 2 =

+2

22 nn

nB

+=

mol%&52 'mol%%+2&%

mol%%+2&%B 2 +

=

%&%%28B 2 =


Tabel A-2 Hasil Penentuan Craksi 0ol 7ukrosa &" +2H22O++)

1un (g) P + n+ n2 B 24 %&5

& %' %&52 '

%&%%+2 %&%%28

44 %&6 %&%%+6 %&%%!6444 %&' %&%%+8 %&%%524V %&8 %&%%2! %&%%65

). "a#$el % Tri*Natriu# P!s+at D!de,ahidrat &Na ) PO - .12' 2O( - NaO'(

Run I


Andri Rizki


6/17

/, '

A0 Tri 9atrium Pos at #ode.ahidrat = +6'% g mol (7.ien.elab& 2%%'.)


OHA0

gram

n 2+ =

9a"lA0

gramn 2 =

g(mol+8&%2g)&)%'

n+ =

g(mol +6'%g%&6

n 2 =

mol52)'&%n+ = mol%%%!&%n 2 =

+2

22 nn

nB

+=

mol%&52 'mol%%%!&%mol%&%%%!

B 2 +=

%%%&%B 2 =


Tabel A-! Hasil Penentuan Craksi 0ol Tri 9atrium Pos at #ode.ahidrat (9a ! PO 5- +2H 2O) 5 9aOH)

1un (g) P + n+ n2 B 24 %&5

& %' %&52 '

%&%%%! %&%%%44 %&6 %&%%%5 %&%%%444 %&' %&%%%5 %&%%+%4V %&8 %&%%%' %&%%+!

B.2.2 Penentuan Penurunan Tekanan a$ Larutan

1. "a#$el % Malt!sa &C 12 ' 22O 11(

Run I

∆ P = P o @ B2

∆ P = 8&8!8' kPa @ %&%%28

= %&22% kPa


Andri Rizki


7/17

/,

#engan rumus tersebut& diperoleh:Tabel A-5 Hasil Penentuan Penurunan Tekanan Dap /arutan 0altosa (" +2H22O++)

1un (g) Po

n+ n2 B 2 EP4 %&5

8&8!8' %&52 '

%&%%+2 %&%%28%&22%44 %&6 %&%%+6 %&%%!6%&2 6444 %&' %&%%+8 %&%%52%&!!++4V %&8 %&%%2! %&%%65%&526

2. "a#$el % "ukr!sa &C 12 ' 22O 11(

Run I

∆ P = P o @ B2

∆ P = 8&8!8' kPa @ %&%%28

= %&22% kPa

#engan rumus tersebut& diperoleh:Tabel A-6 Hasil Penetuan Penurunan Tekanan Dap /arutan 7ukrosa (" +2H22O++)

1un (g) P o n+ n2 B 2 EP4 %&5

8&8!8' %&52 '

%&%%+2 %&%%28%&22%

44 %&6 %&%%+6 %&%%!6%&2 6444 %&' %&%%+8 %&%%52%&!!++4V %&8 %&%%2! %&%%65%&526

). "a#$el % Tri*Natriu# $!s+at d!de,ahidrat &Na ) PO - .12' 2O( - NaO'(

Run I

∆ P = P o @ B2

∆ P = 8&8!8' kPa @ %&%%%

= %&%662 kPa

#engan rumus tersebut& diperoleh:


Andri Rizki


8/17

/, 8

Tabel A-' Hasil Penentuan Penurunan Tekanan Dap /arutan Tri 9atrium Pos at

#ode.ahidrat (9a ! PO 5-+2H2O) 5 9aOH)

1un (g) P o n+ n2 B 2 EP4 %&5

8&8!8' %&52 '

%&%%%! %&%%% %&%66244 %&6 %&%%%5 %&%%% %&% %444 %&' %&%%%5 %&%%+% %&% 884V %&8 %&%%%' %&%%+! %&+%26

B.) Perhitun an Kenaikan Titik Didih Larutan

B.).1 Penentuan M!lalitas Larutan

1. "a#$el % Malt!sa &C 12' 22O 11(

A0 " +2H22O++ = !52&!% g mol (7.ien.elab& 2%+!a)

Run I

onsentrasi larutan awal :

++ P

+%%%@

A0m =

& %'+%%%

@!52&!%%&5

m+ =

+66)&%m+ = m

Volume pelarut setelah mendidih :

V pelarut = V 2 > (V + Vo)

V pelarut = +! > (8&6 >8)

V pelarut = +2&6 ml

0assa pelarut :

P2 = $ pelarut @ V pelarut

P2 = %& '!2! @ +2&6

P2 = +2&%5%5 g


Andri Rizki


9/17

/,

onsentrasi larutan akhir :

22 P

+%%%@A0

m =

+2&%5%5+%%%

@!52&!%

%&5m 2 =

%&% )m 2 = m

#engan rumus diatas& diperoleh:

Tabel A- Hasil Penentuan 0olalitas /arutan 0altosa (" +2H22O++)

1un (g) V pelarut P+ P2 m+ m24 %&5 +2&6

& %'

+2&%5%5 %&+66 %&%44 %&6 +2&6 +2&%5%5 %&+ 5' %&+256444 %&' +5 +!&5856 %&2!!6 %&+!!64V %&8 +5 +!&5856 %&2 86 %&+ %'

2. "a#$el % "ukr!sa &C 12 ' 22O 11(

A0 " +2H22O++ = !52&!% g mol (7.ien.elab& 2%+!b)

Run I


++ P

+%%%@

A0m =

& %'+%%%

@!52&!%

%&5m+ =

+66)&%m+ = m



V pelarut = ++&6 > (8&6 >8)

V pelarut = ++ ml


Andri Rizki


10/17

/, +%

0assa pelarut :


P2 = %& '!2! @ ++

P2 = +%&6 66 g


22 P

+%%%@

A0m =

+%&6 66+%%%

@!52&!%

%&5m 2 =

%&++!!m 2 = m


Tabel A-8 Hasil Penentuan 0olalitas /arutan 7ukrosa (" +2H22O++)

1un (g) V pelarut P+ P2 m+ m24 %&5 ++

& %'

+%&6 66 %&+66 %&++!!44 %&6 +2 ++&% + %&+ 5' %&+!65

444 %&' +2 ++&% + %&2!!6 %&+'264V %&8 +2&6 ++&6688 %&2 86 %&+ 8


A0 Tri 9atrium Pos at #ode.ahidrat (9a ! PO 5-+2H2O) 5 9aOH) = +6'% g mol

(7.ien.elab& 2%%'.)

Run I


++ P

+%%%@A0

m =

& %'+%%%

@+6'%%&6

m+ =

%5+'&%m+ =

m



Andri Rizki


11/17

/, ++


V pelarut = +2 > (8> 8)

V pelarut = +2 m

0assa pelarut :


P2 = %& '!2! @ +2

P2 = ++&6688 g


22 P

+%%%@

A0m =

++&6688+%%%

@+6'%%&6

m 2 =

%26&%m 2 =

m


Tabel A- Hasil Penentuan 0olalitas /arutan Tri 9atrium Pos at #ode.ahidrat(9a ! PO 5-+2H2O) 5 9aOH)

1un (g) V pelarut P+ P2 m+ m24 %&6 +2

& %'

++&6688 %&%5+' %&%2644 %&' +2&6 +2&%5%5 %&%5 %&%!!2444 %& +! +2&62+ %&%682 %&%!4V %& +! +2&62+ %&% 5 %&%5

B.).2 Perhitun an K!nstanta Kenaikan Titik Didih

1. "a#$el % Malt!sa &C 12 ' 22O 11(

Pada T + = %o"& EH


12/17

/, +2

%+%%%!

2''%&++&2' '+&2''%H<

−−

=−

−∆

+%!

+'+&2''%H<

=−∆

8&5+&2''%H< =−∆

&2''5H< =∆kF kg

A0 H 2O = +8&%2 g mol (7.ien.elab& 2%+!d)1 = 8&!+5!5 kF kg-mol- (*eankoplis& + !)

+%%%A0@3H


13/17

/, +!

+!2+&%6!)5&)

%5!)&2i

-m=d

3Tdi

2teori

reg

×

=

=

i = 2&%626

2&%626+6&5+5)

=d

i=d

=d

per.

i per.

=

=

d per. = &6+%2o" m

G+%%&6! 5

6+%2&6! 5&1alatG

G+%%=d

=d=d1alatG

teori

per.teori

×−

=

×−

=

G 1alat = %&!' G

2. "a#$el % "ukr!sa &C 12 ' 22O 11(

Pada T = ! o "&&2''5H< =∆

kF kg (*eankoplis& 2%%!)


+%%%A0@

3H<)1(T=d

2k

teori =

+%%%+8&%2

@2''5&

2 !&+6)!(8&!+5!5)(=d

2

teori

+=

"(m )&6!)5=d teori °=


Andri Rizki


14/17

/, +5

%&++!!+ 8&%

+ 5&226&!=d

mm

3Td3Td=d

i

awalakhir

awalregakhir regi

−

−=

−

−=

d i = 2%& %8o" molal

5%&+ 8%&+'26%&+!65++!!&%

m

nm

m

2

22

+++=

∑=

m 2 = %&+626 m

ETd reg pada m 2 = !&%%%%

+626&%6!)5&)%%%%&!

i

-m=d

3Tdi

2teori

reg

×=

=

i = 2&'% 8

2&'% 82%& %8

=d

i=d

=d

per.

i per.

=

=

d per. = 8&%!65o" m

G+%%&6! 5

%!65&86! 5&1alatG

G+%%=d

=d=d1alatG

teori

per.teori

×−

=

×−

=

G 1alat = '&'+ G



Andri Rizki


15/17

/, +6

Pada T = ! o "&&2''5H< =∆

kF kg (*eankoplis& + !)


+%%%A0

@3H<

)1(T=d

2k

teori =

+%%%+8&%2

@2''5&

2 !&+6)!(8&!+5!5)(=d

2

teori

+=

"(m )&6!)5=d teori °=

%&%26%&%52525&+%88&+'

=d

mm

3Td3Td=d

i

awalakhir

awalregakhir regi

−−

=

−

−=

d i = !%'&'6 + o" molal

5%&%5%&%!%&%!!2%26&%

mn

mm

2

22

+++=

∑=

m 2 = %&%!' m

ETd reg pada m 2 = +!&!2!+

%!')&%6!)5&)!2!+&+!

i

-m=d

3Tdi

teori

reg

×=

=

i = 58&+' 5

58&+' 5!%'&'6 +

=d

i=d

=d

per.

i per.

=

=


Andri Rizki


16/17

/, +'

d per. = '&!''6 o" molal

G+%%&6! 5

!''6&'&6! 51alatG

G+%%=d

=d=d1alatG

teori

per.teori

×−

=

×

−

=

G 1alat = +6&6! G

B.- Perhitun an Td Te!ri

1. "a#$el % Malt!sa &C 12 ' 22O 11(

Run I

2teori m=d3Td ×=

%&%&6! 5Td teori ×=∆

"!+&%3Td oteori =

2. "a#$el % "ukr!sa &C 12 ' 22O 11(

Run I

2teori m=d3Td ×=

%&++!!&6! 53Td teori ×=

"86!&%3Td oteori =


Run I

im=d3Td 2teori ××=

62%&%26&6! 53Td teori ××=

"+6+!+%&Td oteori

=∆


Andri Rizki


17/17

/, +


Andri Rizki

lampiran a b fiizzx

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