perhitungan modul i-vi, tanpa v_kimfis

MODUL ITabung reaksi A Massa

KClO3

Gelas ukur C Volume air dalam

botol B

Volume

O2Sebelum Setelah Sebelum Setelah

25,02 g 25,09 g 100 mg 126,74 g 159,98 g 80 % 33,36 L

24,91 g 24,98 g 100 mg 126,74 g 164,67 g 80 % 38,07 L

25,75 g 25,83 g 100 mg 126,74 g 170,97 g 80 % 44,39 L

24,88 g 24,95 g 100 mg 111,98 g 149,45 g 50 % 37,61 L

24,98 g 25,06 g 100 mg 111,98 g 141,99 g 50 % 30,12 L

25,03 g 25,09 g 100 mg 111,98 g 152,47 g 50 % 40,64 L

T = 27,8oC.

ρ = 996,2738 kg.m-3 = 0,9962738 g.L-1

Perhitungan.

1) Stoikiometri.

a. Berdasarkan berat KClO3.

Reaksi:

2KClO3 2KCl + 3O2

Massa KClO3 = 100 mg = 0,1 g.

Mr = 122,5 g.mol-1

Mol O2:

Mr O2 = 32

Massa O2 = n x Mr

= 9,995x10-7 x 32

Massa O2 = 3,199x10-5 gram.

1

V = n x 22,4

V = 3,199x10-5 x 22,4

V = 2,239x10-5 L.

b. Berdasarkan berat KCl.

Reaksi:

2KClO3 2KCl + 3O2

1. Volume air 80%.

Percobaan 1.

Massa KCl = 0,07 g.

Mr = 74,5 g.mol-1

Mol O2:

Mr O2 = 32

Massa O2 = n x Mr

= 1,409x10-3 x 32

Massa O2 = 0,045 gram.

2

V = n x 22,4

V = 1,409x10-3 x 22,4

V = 0,032 L.

Percobaan 2.

Massa KCl = 0,07 g.

Mr = 74,5 g.mol-1

Mol O2:

Mr O2 = 32

Massa O2 = n x Mr

= 1,409x10-3 x 32

Massa O2 = 0,045 gram.

V = n x 22,4

V = 1,409x10-3 x 22,4

V = 0,032 L.

Percobaan 3.

3

Massa KCl = 0,08 g.

Mr = 74,5 g.mol-1

Mol O2:

Mr O2 = 32

Massa O2 = n x Mr

= 1,611x10-3 x 32

Massa O2 = 0,052 gram.

V = n x 22,4

V = 1,611x10-3 x 22,4

V = 0,036 L.

2. Volume air 50%.

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Percobaan 1.

Massa KCl = 0,07 g.

Mr = 74,5 g.mol-1

Mol O2:

Mr O2 = 32

Massa O2 = n x Mr

= 1,409x10-3 x 32

Massa O2 = 0,045 gram.

V = n x 22,4

V = 1,409x10-3 x 22,4

V = 0,032 L.

Percobaan 2.

Massa KCl = 0,08 g.

Mr = 74,5 g.mol-1

Mol O2:

5

Mr O2 = 32

Massa O2 = n x Mr

= 1,611x10-3 x 32

Massa O2 = 0,052 gram.

V = n x 22,4

V = 1,611x10-3 x 22,4

V = 0,036 L.

Percobaan 3.

Massa KCl = 0,06 g.

Mr = 74,5 g.mol-1

Mol O2:

Mr O2 = 32

Massa O2 = n x Mr

= 1,208x10-3 x 32

Massa O2 = 0,039 gram.

6

V = n x 22,4

V = 1,208x10-3 x 22,4

V = 0,027 L.

2) Persamaan gas nyata.

a) Volume air 80%.

Percobaan 1.

m air = massa gelas ukur sebelum – massa gelas ukur sesudah.

m air = 159,98 – 126,74 = 33,24 gram.

7

Percobaan 2.

m air = massa gelas ukur sebelum – massa gelas ukur sesudah.

m air = 164,67 – 126,74 = 37,93 gram.

8

Percobaan 3.

m air = massa gelas ukur sebelum – massa gelas ukur sesudah.

m air = 170,97 – 126,74 = 44,23 gram.

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b) Volume air 50%.

a. Percobaan 1.

m air = massa gelas ukur sebelum – massa gelas ukur sesudah.

m air = 149,45 – 111,98 = 37,47 gram.

b. Percobaan 2.

m air = massa gelas ukur sebelum – massa gelas ukur sesudah.

m air = 141,99 – 111,98 = 30,01 gram.

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c. Percobaan 3.

m air = massa gelas ukur sebelum – massa gelas ukur sesudah.

m air = 1152,47 – 111,98 = 40,49 gram.

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MODUL II

Massa Volume Kerapatan (ρ)

Piknometer Piknometer

+ zat

Zat

Aquadest 11,009 g 20,921 g 9,912 g 9,915 mL 0,9996974 g/mL

Tween 5% 8,265 g 18,982 g 10,717 g 10,72 mL 0,9997201 g/mL

Tween 10% 8,961 g 21,071 g 12,11 g 12,11 mL 1 g/mL

Tween 15% 11,009 g 18,041 g 7,032 g 7,034 mL 0,9997156 g/mL

Tween 20% 9,158 g 18,084 g 8,926 g 8,928 mL 0,9997759 g/mL

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Suhu

(oC)

Jumlah Tetesan

Aquadest Tween 5% Tween 10% Tween 15% Tween 20%

26,2o 58 tetes 143 tetes 94 tetes 132 tetes 152 tetes

40o 60 tetes 144 tetes 90 tetes 138 tetes 147 tetes

50o 59 tetes 144 tetes 99 tetes 149 tetes 183 tetes

60o 61 tetes 146 tetes 87 tetes 153 tetes 160 tetes

T ruangan = 26,2oC

ρ air literatur = ρ ruangan = 0,9997337 g/mL.

Perhitungan.

Pengenceran.

Tween 20%.

Tween 15%.

Tween 10%.

Tween 5%.

Kerapatan (ρ).

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ρ literatur = 0,9997337 g/mL

Tween 5%.

Tween 10%.

Tween 15%.

Tween 20%.

Tegangan permukaan.

Rumus:

Tween 5%.

T = 26,2oC.

γo = 726 N/m

T = 40oC.

γo = 701 N/m

T = 50oC.

γo = 682 N/m

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T = 60oC.

γo = 668 N/m

Tween 10%.

T = 26,2oC.

γo = 726 N/m

T = 40oC.

γo = 701 N/m

T = 50oC.

γo = 682 N/m

T = 60oC.

γo = 668 N/m

Tween 15%.

T = 26,2oC.

γo = 726 N/m

T = 40oC.

γo = 701 N/m

15

T = 50oC.

γo = 682 N/m

T = 60oC.

γo = 668 N/m

Tween 20%.

T = 26,2oC.

γo = 726 N/m

T = 40oC.

γo = 701 N/m

T = 50oC.

γo = 682 N/m

T = 60oC.

γo = 668 N/m

MODUL III

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ANALYSIS DATA

Sample 30o 40o

Aquades 29,02” 24,92” 25,14” 28,15” 27,54” 29,02”

Gliserin 5% 38,80” 37,10” 37,72” 40,02” 42,66” 38,86”

Gliserin 10% 46,09” 44,68” 45,64” 52,87” 52,35” 55,14”

Gliserin 15% 1’36,85” 1’33,41” 1’39,47” 1’46,32” 1’46,06” 1’48,11”

Sample 50o

Aquades 25,41” 22,61” 23,02”

Gliserin 5% 34,05” 34,35” 34,76”

Gliserin 10% 39,29” 42,09” 41,49”

Gliserin 15% 1’20,08” 1’22,06” 1’46,32”

o Calculate of solution

Sample Weight (p +

s)

Weight of

sample

Weight of

piknometre

Aquades 35,08 g 23,77 g 11,31 g 0,9508

Gliserin 5% 35,47 g 24,14 g 11,33 g 0,9656

Gliserin 10% 35,72 g 24,39 g 11,33 g 0,9756

Gliserin 15% 36,64 g 24,34 g 12,30 g 0,9736

Literature and

270 400 500

0,894 0,656 0,549

0,997 0,992 0,992

Calculation of viscosity

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The rate of the time solution to reach distance from M to N

Sample 300 400 500

Aquades 22,82” 23,24” 25,03”

Gliserin 5% 34,39” 40,51” 37,41”

Gliserin 10% 41,79” 52,61” 45,47”

Gliserin 15% 1’21,07” 1’46,19” 1’38,16”

For 300

Gliserin 5%

gliserin 5% = 1,495

Gliserin 10%

gliserin 10% = 1,98

Gliserin 15%

gliserin 15% = 3,989

For 400

Gliserin 5%

gliserin 5% = 0,9544

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Gliserin 10%

gliserin 10% = 1,172

Gliserin 15%

gliserin 15% = 2,525

For 500

Gliserin 5%

gliserin 5% = 0,81

Gliserin 10%

gliserin 10% = 0,99

Gliserin 15%

gliserin 15% = 1,91

MODUL IV

1) Variation concentration.

No. V KI

0,4 N

V K2S2O8

0,02 N

V Na2S2O3

0,1 N

Aquades Amylum tf f

1. 20 mL 10 mL 1 mL 15 mL 10 drops 261 0,5 0,30

2. 20 mL 15 mL 1 mL 10 mL 10 drops 145 0,34 0,18

3. 20 mL 20 mL 1 mL 5 mL 10 drops 115 0,25 0,12

4. 20 mL 25 mL 1 mL - 10 drops 104 0,20 0,09

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Calculation:

a)

b)

c)

d)

No.X =

Y = tf X.Y X2

1. 0,30 261 78,3 0,090

2. 0,18 145 26,1 0,0324

3. 0,12 115 13,8 0,0144

4. 0,09 104 9,36 0,0081

∑ 0,69 625 127,56 0,1449

Y = mX + C

+ C

20

m = at

C = bt

The line equation is:

Y = 763,19X + 24,6

a) Ea = (2,303 R.T.log tf) + k

Ea = (2,303 8,314.301.log 261) + 0,003

Ea = 13927,80 kal/mole.

b) Ea = (2,303 R.T.log tf) + k

Ea = (2,303 8,314.301.log 145) + 0,003

Ea = 12456,59 kal/mole.

c) Ea = (2,303 R.T.log tf) + k

Ea = (2,303 8,314.301.log 115) + 0,003

Ea = 11876,40 kal/mole.

d) Ea = (2,303 R.T.log tf) + k

Ea = (2,303 8,314.301.log 104) + 0,003

Ea = 11624,75 kal/mole.

The Influence by concentrate:

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a) V1.M1 = V2.M2

20.0,4 = 10.M2

M2 = 0,8 M.

b) V1.M1 = V2.M2

20.0,4 = 15.M2

M2 = 0,54 M.

c) V1.M1 = V2.M2

20.0,4 = 20.M2

M2 = 0,4 M.

d) V1.M1 = V2.M2

20.0,4 = 25.M2

M2 = 0,32 M.

2) Variation temperature.

No. V KI

0,4 N

V K2S2O8

0,02 N

V Na2S2O3

0,1 N

Amylum T (oC) T (oK) tf log tf 1/T

(K-1)

1. 20 mL 20 mL 1 Ml 10 drops 40 313 124 2,09 0,0032

2. 20 mL 20 mL 1 mL 10 drops 50 323 30 1,48 0,0031

3. 20 mL 20 mL 1 mL 10 drops 60 333 19 1,28 0,0030

4. 20 mL 20 mL 1 mL 10 drops 68 341 20 1,30 0,0029

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Calculation:

a) T = 40oC.

b) T = 50oC.

c) T = 60oC.

d) T = 68oC.

No. X = 1/T Y = log tf X.Y X2

1. 3,2 x 10-3 2,09 6,69 x 10-3 1,024 x 10-5

2. 3,1 x 10-3 1,48 4,59 x 10-3 9,61 x 10-6

3. 3,0 x 10-3 1,28 3,84 x 10-3 9,00 x 10-6

4. 2,9 x 10-3 1,30 3,77 x 10-3 8,41 x 10-6

∑ 1,22 x 10-2 6,15 1,89 x 10-2 3,726 x 10-5

Y = mX + C

+ C

23

m = at

C = bt

The line equation is:

Y = 2570X – 6,301

a) T = 40oC.

b) T = 50oC.

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c) T = 60oC.

d) T = 68oC.

The influence by temperature:

a) T = 40oC.

b) T = 50oC.

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c) T = 60oC.

d) T = 68oC.

MODUL VI

II. Experiment Result.

1. Standardization NaOH by H2C2O4.

m = 0,25 gram.

BE = 63,035.

V = 20 mL.

Titration:

V1 (NaOH) = 8,35 mL.

(NaOH) N1.V1 = N2.V2 (H2C2O4)

8,35 mL. N1 = 0,1983 N. 20 mL

N1 = 0,4749 N.

q = 390 calorie H = - 7800 calorie/ equivalent.

V2 (NaOH) = 8,2 mL. N (NaOH) = 0,483 N.

V3 (NaOH) = 8,1 mL. N (NaOH) = 0,489 N.

V4 (NaOH) = 8 mL. N (NaOH) = 0,495 N.

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2. Heat capacity equipment.

T1 (oC) 28

T2 (oC) 41

T3 (oC) 33

W 30

50 (T2 – T3) = W (T3 – T1) + 50 (T3 – T1).

50 ( 314 – 306) = W (306 – 301) + 50 (306 – 301)

50.8 = 5W + 50.5

5W = 400 – 250

5W = 150

W = 30.

3. Heat neutralization.

a. Heat neutralization NaOH – HCl.

Time Themperature (oC)

5 s 30

10 s 31

15 s 31

20 s 31

30 s 30,5

60 s 31

2 minute 29

3 minute 30

5 minute 30

7 minute 30

t4 = 28 oC, t5 = 31 oC.

q = 100 (T5 – T4) + W (T5 – T4)

q = 100 (304 – 301) + 30 (304 – 301).

q = 100.3 + 30.3

q = 300 + 90

q = 3900 calorie

H = - 20.q

27

H = - 20.390

H = - 7.800 calorie/ equivalent.

b. Heat neutralization NaOH – HNO3.

Time Themperature (oC)

5 s 29

10 s 29,5

15 s 29,5

20 s 29,5

30 s 29,5

60 s 29,5

2 minute 29,2

3 minute 29,2

5 minute 29,1

7 minute 29,1

t4 = 27 oC, t5 = 29,5 oC.

q = 100 (T5 – T4) + W (T5 – T4)

q = 100 (302,5 – 300) + 30 (302,5 – 300).

q = 100.2,5 + 30.2,5

q = 250 + 75

q = 325 calorie

H = - 20.q

H = - 20.325

H = - 6.500 calorie/ equivalent.

c. Heat neutralization NaOH – CH3COOH.

Time Themperature (oC)

5 s 30

10 s 30

15 s 30

20 s 30

30 s 30

28

60 s 30

2 minute 29,9

3 minute 29,9

5 minute 29,9

7 minute 29,9

t4 = 27 oC, t5 = 30 oC.

q = 100 (T5 – T4) + W (T5 – T4)

q = 100 (303 – 300) + 30 (303 – 300).

q = 100.3 + 30.3

q = 300 + 90

q = 390 calorie

H = - 20.q

H = - 20.390

H = - 7.800 calorie/ equivalent.

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