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Lampiran Perhitungan a. Pengenceran bahan a. 0,5 M M 1 .V 1 =M 2 .V 2 1,00 M. V 1 = 0,50 M. 100 mL 1,00 V 1 = 50 mL V 1 = 50 mL b. 0,1 M M 1 .V 1 =M 2 .V 2 1,00 M. V 1 = 0,10 M. 100 mL 1,00 V 1 = 10 mL V 1 = 10 mL c. 0,05 M M 1 .V 1 =M 2 .V 2 1,00 M. V 1 = 0,05 M. 100 mL 1,00 V 1 = 5 mL V 1 = 5 mL d. 0,01 M M 1 .V 1 =M 2 .V 2 1,00 M. V 1 = 0,01 M. 100 mL 1,00 M. V 1 = 1 mL V 1 = 1 mL b. Daya Hantar Listrik Berbagai Senyawa 1. Minyak tanah V = I R R = 5 V 0,0 A = 0 Ω L = 1 R = 1 0 Ω = 0 Ω -1 2. Asam cuka glasial V = I R R = 5 V 0,0 A = 0 Ω L = 1 R = 1 0 Ω = 0 Ω -1 3. Akuades V = I R R = 5 V 0,1 x 10 3 A = 5 x 10 4 Ω L = 1 R = 1 5 x 10 4 Ω = 0,2 x 10 -4 Ω -1 4. NaCl V = I R R = 5 V 3,0 x 10 3 A = 1,67 x 10 3 Ω L = 1 R = 1 1,67 x 10 3 Ω = 6 x 10 -4 Ω -1

Lampiran Perhitungan Kimfis 6

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Page 1: Lampiran Perhitungan Kimfis 6

Lampiran Perhitungan

a. Pengenceran bahan

a. 0,5 M

M1. V1 = M2. V2

1,00 M. V1 = 0,50 M. 100 mL

1,00 V1 = 50 mL

V1 = 50 mL

b. 0,1 M

M1. V1 = M2. V2

1,00 M. V1 = 0,10 M. 100 mL

1,00 V1 = 10 mL

V1 = 10 mL

c. 0,05 M

M1. V1 = M2. V2

1,00 M. V1 = 0,05 M. 100 mL

1,00 V1 = 5 mL

V1 = 5 mL

d. 0,01 M

M1. V1 = M2. V2

1,00 M. V1 = 0,01 M. 100 mL

1,00 M. V1 = 1 mL

V1 = 1 mL

b. Daya Hantar Listrik Berbagai Senyawa

1. Minyak tanah

V = I R

R = 5 V

0,0 A

= 0 Ω

L = 1

R

= 1

0 Ω

= 0 Ω-1

2. Asam cuka glasial

V = I R

R = 5 V

0,0 A

= 0 Ω

L = 1

R

= 1

0 Ω

= 0 Ω-1

3. Akuades

V = I R

R = 5 V

0,1 x 10−3A

= 5 x 104 Ω

L = 1

R

= 1

5 x 104Ω

= 0,2 x 10-4

Ω-1

4. NaCl

V = I R

R = 5 V

3,0 x 10−3A

= 1,67 x 103 Ω

L = 1

R

= 1

1,67 x 103Ω

= 6 x 10-4

Ω-1

Page 2: Lampiran Perhitungan Kimfis 6

c. Pengaruh konsentrasi terhadap Daya Hantar Listrik larutan elektrolit

1. Kelompok 1

CH3COOH

0,01 M

V = I R

R = 5 V

1,5 x 10−3A

= 3,33 x 103 Ω

L = 1

R

= 1

3,33 x 103Ω

= 3 x 10-4

Ω-1

0,05 M

V = I R

R = 5 V

1,6 x 10−3A

= 3,125 x 103 Ω

L = 1

R

= 1

3,125 x 103Ω

= 3,2 x 10-4

Ω-1

0,1 M

V = I R

R = 5 V

2,7 x 10−3A

= 1,85 x 103 Ω

L = 1

R

= 1

1,85 x 103Ω

= 5,4 x 10-4

Ω-1

0,5 M

V = I R

R = 5 V

2,8 x 10−3A

= 1,8 x 103 Ω

L = 1

R

= 1

1,8 x 103Ω

= 5,6 x 10-4

Ω-1

1,0 M

V = I R

R = 5 V

3,4 x 10−3A

Page 3: Lampiran Perhitungan Kimfis 6

= 1,5 x 103 Ω

L = 1

R

= 1

1,5 x 103Ω

= 6,7 x 10-4

Ω-1

- NH4OH

0,01 M

V = I R

R = 5 V

2,3 x 10−3A

= 2,2 x 103 Ω

L = 1

R

= 1

2,2 x 103Ω

= 4,5 x 10-4

Ω-1

0,05 M

V = I R

R = 5 V

3,0 x 10−3A

= 1,67 x 103 Ω

L = 1

R

= 1

1,67 x 103Ω

= 6 x 10-4

Ω-1

0,1 M

V = I R

R = 5 V

3,2 x 10−3A

= 1,5625 x 103 Ω

L = 1

R

= 1

1,5625 x 103Ω

= 6,4 x 10-4

Ω-1

0,5 M

V = I R

R = 5 V

3,4 x 10−3A

= 1,5 x 103 Ω

L = 1

R

= 1

1,5 x 103Ω

= 6,7 x 10-4

Ω-1

1,0 M

Page 4: Lampiran Perhitungan Kimfis 6

V = I R

R = 5 V

3,7 x 10−3A

= 1,35 x 103 Ω

L = 1

R

= 1

1,35 x 103Ω

= 7,4 x 10-4

Ω-1

- HCl

0,01 M

V = I R

R = 5 V

1,8 x 10−3A

= 2,78 x 103 Ω

L = 1

R

= 1

2,78 x 103Ω

= 3,6 x 10-4

Ω-1

0,05 M

V = I R

R = 5 V

3,3 x 10−3A

= 1,5 x 103 Ω

L = 1

R

= 1

1,5 x 103Ω

= 6,7 x 10-4

Ω-1

0,1 M

V = I R

R = 5 V

3,7 x 10−3A

= 1,35 x 103 Ω

L = 1

R

= 1

1,35 x 103Ω

= 7,4 x 10-4

Ω-1

0,5 M

V = I R

R = 5 V

3,8 x 10−3A

= 1,32 x 103 Ω

L = 1

R

= 1

1,32 x 103Ω

Page 5: Lampiran Perhitungan Kimfis 6

= 7,6 x 10-4

Ω-1

1,0 M

V = I R

R = 5 V

3,9 x 10−3A

= 1,28 x 103 Ω

L = 1

R

= 1

1,28 x 103Ω

= 7,8 x 10-4

Ω-1

- NaOH

0,01 M

V = I R

R = 5 V

2,8 x 10−3A

= 1,78 x 103 Ω

L = 1

R

= 1

1,78 x 103Ω

= 5,6 x 10-4

Ω-1

0,05 M

V = I R

R = 5 V

3,2 x 10−3A

= 1,5625 x 103 Ω

L = 1

R

= 1

1,5625 x 103Ω

= 6,4 x 10-4

Ω-1

0,1 M

V = I R

R = 5 V

3,3x 10−3A

= 1,52 x 103 Ω

L = 1

R

= 1

1,52 x 103Ω

= 6,6 x 10-4

Ω-1

0,5 M

V = I R

R = 5 V

3,5 x 10−3A

= 1,43 x 103 Ω

Page 6: Lampiran Perhitungan Kimfis 6

L = 1

R

= 1

1,43 x 103Ω

= 7 x 10-4

Ω-1

1,0 M

V = I R

R = 5 V

3,8 x 10−3A

= 1,32 x 103 Ω

L = 1

R

= 1

1,32 x 103Ω

= 7,6 x 10-4

Ω-1

2. Kelompok 2

- NaCl

0,01 M

V = I R

R = 5 V

2 x 10−3A

= 2,5 x 103 Ω

L = 1

R

= 1

2,5 x 103Ω

= 4 x 10-4

Ω-1

0,05 M

V = I R

R = 5 V

2,3 x 10−3A

= 2,17 x 103 Ω

L = 1

R

= 1

2,17 x 103Ω

= 4,6 x 10-4

Ω-1

0,1 M

V = I R

R = 5 V

2,6 x 10−3A

= 1,92 x 103 Ω

L = 1

R

= 1

1,92 x 103Ω

= 5,2 x 10-4

Ω-1

0,5 M

Page 7: Lampiran Perhitungan Kimfis 6

V = I R

R = 5 V

3,4 x 10−3A

= 1,47 x 103 Ω

L = 1

R

= 1

1,47 x 103Ω

= 6,8 x 10-4

Ω-1

1,0 M

V = I R

R = 5 V

3,6 x 10−3A

= 1,39 x 103 Ω

L = 1

R

= 1

1,39 x 103Ω

= 7,2 x 10-4

Ω-1

- NaBr

0,01 M

V = I R

R = 5 V

1,2 x 10−3A

= 4,17 x 103 Ω

L = 1

R

= 1

4,17 x 103Ω

= 2,4 x 10-4

Ω-1

0,05 M

V = I R

R = 5 V

1,7 x 10−3A

= 2,9 x 103 Ω

L = 1

R

= 1

2,9 x 103Ω

= 3,4 x 10-4

Ω-1

0,1 M

V = I R

R = 5 V

1,8 x 10−3A

= 2,78 x 103 Ω

L = 1

R

= 1

2,78 x 103Ω

Page 8: Lampiran Perhitungan Kimfis 6

= 3,6 x 10-4

Ω-1

0,5 M

V = I R

R = 5 V

2,0 x 10−3A

= 2,5 x 103 Ω

L = 1

R

= 1

2,5 x 103Ω

= 4,0 x 10-4

Ω-1

1,0 M

V = I R

R = 5 V

3,7 x 10−3A

= 1,35 x 103 Ω

L = 1

R

= 1

1,35 x 103Ω

= 7,4 x 10-4

Ω-1

- NH4Cl

0,01 M

V = I R

R = 5 V

3 x 10−3A

= 1,67 x 103 Ω

L = 1

R

= 1

1,67 x 103Ω

= 6,0 x 10-4

Ω-1

0,05 M

V = I R

R = 5 V

4,1 x 10−3A

= 1,22 x 103 Ω

L = 1

R

= 1

1,22 x 103Ω

= 8,2 x 10-4

Ω-1

0,1 M

V = I R

R = 5 V

4,3 x 10−3A

= 1,16 x 103 Ω

Page 9: Lampiran Perhitungan Kimfis 6

L = 1

R

= 1

1,16 x 103Ω

= 8,6 x 10-4

Ω-1

0,5 M

V = I R

R = 5 V

4,5 x 10−3A

= 1,1 x 103 Ω

L = 1

R

= 1

1,1 x 103Ω

= 9,0 x 10-4

Ω-1

1,0 M

V = I R

R = 5 V

5,4 x 10−3A

= 0,93 x 103 Ω

L = 1

R

= 1

0,93 x 103Ω

= 10,7 x 10-4

Ω-1