# Penilaian Akhir Tahun 2016-2017 / Kimia Xi IPA

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Kelas   : XI IPATanggal: 30 Mei 2017Durasi  : 120 menitperhatikan semua perintah soal dan pilihan Jawaban."SELAMAT MENGERJAKAN"

• 1.

• 2.

### Campuran berikut yang dapat membentuk larutan penyangga adalah ….

• A.

100 mL NaOH 0,1 M + 100 mL HCl 0,1 M

• B.

100 mL NaOH 0,1 M + 100 mL NaCN 0,1 M

• C.

100 mL NaCN 0,1 M + 100 mL HCN 0,1 M

• D.

100 mL NH4OH 0,1 + 50 mL H2SO4 0,1 M

• E.

100 mL K2SO4 0,1 M + 50 mL H2SO4 0,1 M

C. 100 mL NaCN 0,1 M + 100 mL HCN 0,1 M
Explanation
IPK

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• 3.

### Untuk membentuk larutan penyangga dengan pH =9 maka 100 mL larutan HCl 0,1 M harus dicampur dengan larutan NH4OH 0,2 M sebanyak …. ( Kb NH4OH = 10-5 )

• A.

100 mL

• B.

150 mL

• C.

200 mL

• D.

250 mL

• E.

300 mL

A. 100 mL
Explanation
To form a buffer solution with a pH of 9, the concentration of the weak base (NH4OH) should be equal to the concentration of the strong acid (HCl). Since the concentration of HCl is given as 0.1 M, we need to find the volume of NH4OH solution that contains the same amount of moles. By using the equation C1V1 = C2V2, where C1 and V1 are the concentration and volume of the HCl solution, and C2 and V2 are the concentration and volume of the NH4OH solution, we can calculate that the volume of NH4OH solution needed is 100 mL.

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• 4.

### Jika larutan CH3COOH dan CH3COONa dengan konsentrasi yang sama dicampur untuk membentuk larutan penyangga dengan pH = 6 – log 5, maka perbandingan volume yang harus dicampurkan adalah …. ( Ka CH3COOH = 10-5 )

• A.

1:1

• B.

1:2

• C.

1:3

• D.

2:1

• E.

2:3

B. 1:2
Explanation
The correct answer is 1:2 because a buffer solution is formed when a weak acid and its conjugate base are mixed in equal amounts. In this case, CH3COOH is a weak acid and CH3COONa is its conjugate base. The pH of the buffer solution is given as 6 - log 5, which indicates that the concentration of H+ ions is 10^-6 M. The Ka value of CH3COOH is given as 10^-5, which means that the concentration of CH3COO- ions is also 10^-5 M. To achieve equal concentrations of CH3COOH and CH3COO- ions in the buffer solution, the volume ratio of CH3COOH to CH3COONa should be 1:2.

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• 5.

### Di antara campuran berikut, yang dapat membentuk larutan penyangga adalah ….

• A.

HCl dan NH4Cl

• B.

NaOH dan CH3COONa

• C.

HBr dan NaBr

• D.

KOH dan KCl

• E.

NH4OH dan NH4Cl

E. NH4OH dan NH4Cl
Explanation
During the night the high contrast between the bright moon and the night's dark skies make the Moon look white.

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• 6.

### Sejumlah NaOH 1 M diperlukan untuk membuat 150 mL larutan penyangga dengan CH3COOH 1 M. Agar diperoleh pH = 5 – log 2, maka volume NaOH tersebut adalah …. ( Ka CH3COOH 1 M = 10-5 )

• A.

25 mL

• B.

37,5 mL

• C.

55 mL

• D.

67,5 mL

• E.

100 mL

B. 37,5 mL
Explanation
To calculate the volume of NaOH needed to achieve a specific pH, we can use the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). In this case, the pKa is 5 (since it is given as -log 10^-5), and we want the pH to be 5 - log 2. By substituting these values into the equation and solving for [A-]/[HA], we find that [A-]/[HA] = 2. Since the concentration of CH3COOH is 1 M, this means that the concentration of CH3COO- should be 2 M.

To calculate the volume of NaOH needed, we can use the equation M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. Substituting the values, we find that (1 M)(V1) = (2 M)(150 mL), which gives us V1 = 75 mL. However, since we are adding NaOH, we need to subtract the initial volume of CH3COOH from this amount. Therefore, the volume of NaOH needed is 75 mL - 37.5 mL = 37.5 mL.

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• 7.

### Untuk mengubah 100 mL larutan HCl dengan pH = 2 menjadi larutan dengan pH = 9, diperlukan NH4OH 0,01 M sebanyak …. ( Kb NH4OH = 10-5 )

• A.

10 mL

• B.

50 mL

• C.

100 mL

• D.

150 mL

• E.

200 mL

C. 100 mL
Explanation
To change the pH of a solution from 2 to 9, we need to add a base to neutralize the acidity of the solution. In this case, NH4OH is used as the base. The concentration of NH4OH is given as 0.01 M and its Kb value is given as 10^-5. The volume of NH4OH needed can be calculated using the equation for the concentration of OH- ions in a weak base solution. By rearranging the equation, we can find that the volume of NH4OH needed is equal to the concentration of NH4OH multiplied by the volume of the HCl solution divided by the concentration of OH- ions. Plugging in the given values, we find that the volume of NH4OH needed is 100 mL.

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• 8.

### Untuk mengubah 100 mL larutan CH3COOH 0,1 M yang pH-nya 3 agar menjadi 6 diperlukan larutan NaOH 0,1 M sebanyak ….

• A.

10 mL

• B.

55 mL

• C.

100 mL

• D.

210 mL

• E.

1.100 mL

C. 100 mL
Explanation
To change the pH of a 100 mL solution of CH3COOH from 3 to 6, it is necessary to add 100 mL of a 0.1 M NaOH solution. The pH of a solution is determined by the concentration of hydrogen ions (H+). Adding a base such as NaOH will increase the concentration of hydroxide ions (OH-) in the solution, which will react with the hydrogen ions to form water. This will result in a decrease in the concentration of H+ ions and an increase in the pH of the solution. The volume of NaOH needed to achieve this change in pH depends on the initial concentration of CH3COOH and the desired final pH. In this case, adding 100 mL of a 0.1 M NaOH solution is sufficient to achieve the desired pH of 6.

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• 9.

### Di antara larutan berikut yang nilai pH-nya relative tetap jika diencerkan atau ditambah sedikit asam atau basa adalah ….

• A.

NaCl dan HCl

• B.

H2SO4 dan K2SO4

• C.

NH2 dan NH4Br

• D.

NaOH dan HNO3

• E.

NH4OH dan NaOH

C. NH2 dan NH4Br
Explanation
The correct answer is NH2 dan NH4Br. This is because NH2 is a weak base and NH4Br is its conjugate acid. When NH2 is diluted or a small amount of acid or base is added, the equilibrium between NH2 and NH4Br will shift slightly, but the pH of the solution will remain relatively constant. This is due to the presence of the conjugate acid-base pair, which acts as a buffer and helps to maintain the pH.

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• 10.

### Massa CH3COOK ( Mr = 98 ) yang harus ditambahkan ke dalam 200 mL larutan CH3COOH 0,2 M agar diperoleh larutan penyangga denga pH = 5 adalah …..

• A.

2,45 gram

• B.

3,92 gram

• C.

12,15 gram

• D.

24,50 gram

• E.

39,2 gram

B. 3,92 gram
Explanation
To prepare a buffer solution with a pH of 5, we need to use a weak acid and its conjugate base. In this case, the weak acid is CH3COOH (acetic acid) and its conjugate base is CH3COO- (acetate ion). The Henderson-Hasselbalch equation can be used to calculate the ratio of the concentrations of the weak acid and its conjugate base in the buffer solution. By plugging in the given pH of 5 and the pKa value of acetic acid (4.76), we can solve for the ratio [CH3COO-]/[CH3COOH]. Once we have the ratio, we can calculate the amount of CH3COOK (potassium acetate) needed to achieve the desired concentration of the acetate ion in the buffer solution. The correct answer is 3.92 grams.

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• 11.

### Untuk membentuk larutan penyangga dengan pH = 5 maka 100 mL larutan HCN 0,2 M harus dicampur dengan larutan NaOH 0,1 M sebanyak …. ( Ka HCN = 10-5 )

• A.

100 mL

• B.

150 mL

• C.

200 mL

• D.

250 mL

• E.

300 mL

A. 100 mL
Explanation
To form a buffer solution with a pH of 5, the concentration of the acid and its conjugate base should be equal. In this case, the acid is HCN and its conjugate base is CN-. The Ka value for HCN is given as 10^-5. To calculate the amount of NaOH solution needed, we can use the equation for the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Since the pH is given as 5 and pKa is -log(Ka) = -log(10^-5) = 5, we can substitute these values into the equation and solve for the ratio [A-]/[HA]. The ratio [A-]/[HA] is equal to the ratio of the volumes of the two solutions, so we can set up the equation (100 mL)/(100 mL + x) = 1, where x is the volume of the NaOH solution needed. Solving for x, we find that x = 100 mL. Therefore, 100 mL of NaOH solution is needed to form a buffer solution with a pH of 5.

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• 12.

### Pernyataan tentang larutan penyangga yang paling tepat adalah ….

• A.

pH larutan penyangga tidak mungkin berubah oleh penambahan asan dan basa

• B.

Larutan penyangga dapat dibuat dari campuran garam dengan basa kuat berlebihan

• C.

Larutan penyangga dapat dibuat dari larutan basa dengan asam kuat berlebihan

• D.

Larutan penyangga dapat dibuat dari campuran larutan asam lemah dengan basa konjugasinya

• E.

Larutan penyangga dapat dibuat dari campuran asam kuat berlebih dengan basa lemah

D. Larutan penyangga dapat dibuat dari campuran larutan asam lemah dengan basa konjugasinya
Explanation
A buffer solution is made by mixing a weak acid with its conjugate base or a weak base with its conjugate acid. This combination allows the buffer solution to resist changes in pH when small amounts of acid or base are added. Therefore, the statement "larutan penyangga dapat dibuat dari campuran larutan asam lemah dengan basa konjugasinya" is the most accurate statement about buffer solutions.

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• 13.

### Sistem larutan penyangga yang bekerja untuk mempertahankan nilai pH cairan protoplasma sel adalah ….

• A.

HPO4^2- / PO4^3-

• B.

H2PO4^- / HPO4^2-

• C.

HCO3^- / CO3^2-

• D.

H2CO3 / HCO3^2-

• E.

CH3COOH / CH3COO^-

B. H2PO4^- / HPO4^2-
Explanation
The correct answer is H2PO4^- / HPO4^2-. This is because the H2PO4^- / HPO4^2- buffer system is able to maintain the pH of the protoplasmic fluid within a narrow range. In this system, H2PO4^- acts as a weak acid and can donate a proton to maintain the pH when it becomes too basic. On the other hand, HPO4^2- acts as a weak base and can accept a proton to maintain the pH when it becomes too acidic. Together, they work to resist changes in pH and keep the protoplasmic fluid stable.

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• 14.

### Dicampurkan larutan HNO2 dengan larutan NaOH membentuk larutan penyangga. Setelah reaksi terdapat 0,02 mol NaNO2 dan 0,47 gram HNO2. pH larutan penyangga tersebut adalah …. ( Ka HNO2 = 4 x 10-4; Mr HNO2 = 47 )

• A.

4 â€“ log 2

• B.

4 â€“ log 4

• C.

4 â€“ log 8

• D.

8 + log 4

• E.

8 + log 2

A. 4 â€“ log 2
Explanation
The given question involves the formation of a buffer solution by mixing HNO2 and NaOH. The reaction results in the formation of NaNO2 and H2O. The concentration of NaNO2 can be determined by the number of moles given, which is 0.02 mol. The concentration of HNO2 can be calculated using the given mass of 0.47 grams and the molar mass of HNO2, which is 47 g/mol. By using the Henderson-Hasselbalch equation, pH = pKa + log ([A-]/[HA]), where [A-] is the concentration of the conjugate base (NaNO2) and [HA] is the concentration of the acid (HNO2), we can calculate the pH. The pKa value for HNO2 is given as 4 x 10^-4. By substituting the values into the equation, we get pH = 4 - log 2.

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• 15.

### Sebanyak 0,1 mol NH4OH ( Kb = 10-5 ) dicampurkan dengan 0,05 mol NH4Cl hingga volume 1 liter. pH larutan yang dihasilkan adalah ….

• A.

5

• B.

5 + log 2

• C.

5 â€“ log 2

• D.

9 + log 2

• E.

9 â€“ log 2

D. 9 + log 2
Explanation
The correct answer is 9 + log 2. This is because when NH4OH and NH4Cl are mixed together, they undergo a reaction to form NH3 and HCl. NH3 is a weak base and HCl is a strong acid. The NH3 will react with water to produce NH4+ and OH-. Since NH4OH is a weak base, it will not fully dissociate in water and therefore OH- concentration will be less than NH4+ concentration. This will result in a basic solution. The pH of a basic solution is calculated using the equation pH = 14 - pOH. pOH is calculated using the equation pOH = -log[OH-]. Since OH- concentration is less than NH4+ concentration, pOH will be greater than 5. Therefore, the pH of the solution will be greater than 9. Adding log 2 to 9 will give a value greater than 9, which matches the given answer.

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• 16.

### Sebanyak 25 mL etilamina 0,1 M ( C2H5NH2, Kb = 5,6 x 10-4 ) dititrasi dengan HCl 0,1 M. Pernyataan yang benar untuk hasil titrasi setelah penambahan 10 mL HCl adalah ….

• A.

[H^+] < 10^-7

• B.

Terbentuk larutan penyangga yang bersifat asam

• C.

[C2H5NH2] > [C2H5NH3^+]

• D.

[OH^-] = [C2H5NH3^+]

• E.

Jumlah ion Cl^- < 0,1 mmol

C. [C2H5NH2] > [C2H5NH3^+]
Explanation
The correct answer is [C2H5NH2] > [C2H5NH3+]. This is because ethylamine (C2H5NH2) is a weak base and HCl is a strong acid. When HCl is added to ethylamine, it will react to form ethylammonium ion (C2H5NH3+). Since ethylamine is a weak base, it will not completely ionize, resulting in a higher concentration of ethylamine compared to ethylammonium ion in the solution. Therefore, [C2H5NH2] > [C2H5NH3+] is the correct statement.

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• 17.

### Rumusan hasil kali kelarutan (Ksp) Ag2CrO4 dinyatakan sebagai ….

• A.

[Ag] [CrO4]

• B.

[Ag^+] [CrO4^2-]

• C.

[Ag^+]^2 [CrO4^2-]

• D.

[Ag^+]^2 [CrO4^-]^4

• E.

[Ag^4+]^2 [CrO2]^4

C. [Ag^+]^2 [CrO4^2-]
Explanation
The correct answer is [Ag^+]^2 [CrO4^2-]. This is because the expression for the solubility product constant (Ksp) of Ag2CrO4 is given by the product of the concentrations of Ag+ ions and CrO4^2- ions raised to their respective stoichiometric coefficients in the balanced equation. In this case, the balanced equation is 2Ag+ + CrO4^2- = Ag2CrO4, which shows that the stoichiometric coefficients for Ag+ and CrO4^2- are both 2. Therefore, the correct expression for Ksp is [Ag^+]^2 [CrO4^2-].

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• 18.

### Jika kelarutan Ca(OH)2 dalam air adalah s mol/L, hasil kali kelarutan Ca(OH)2 adalah ….

• A.

S^2

• B.

S^3

• C.

2s^3

• D.

4s^3

• E.

16s^4

D. 4s^3
Explanation
The given question is asking for the result of the solubility product of Ca(OH)2 in water, which is represented by the expression 4s^3. In a saturated solution of Ca(OH)2, the concentration of Ca2+ ions and OH- ions are both equal to "s". Since the solubility product expression for Ca(OH)2 is [Ca2+][OH-]^2, plugging in the concentration of "s" for both ions gives us 4s^3 as the result.

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• 19.

### Diketahui Ksp AgCl = 1,6 x 10-10. Satu mol AgCl dimasukkan ke dalam satu liter air, maka dalam larutan terdapat ….

• A.

Konsentrasi AgCl = 1,6 x 10^-10 M

• B.

Konsentrasi ion Ag^+ = 1,6 x 10^-10 M

• C.

Konsentrasi ion Cl^- = 1,6 x 10^-10 M

• D.

[Ag^+] + [Cl^-] = 1,6 x 10^-10

• E.

[Ag^+] [C^l-] = 1,6 x 10^-10

E. [Ag^+] [C^l-] = 1,6 x 10^-10
Explanation
The given answer states that the product of the concentrations of Ag+ and Cl- ions in the solution is equal to 1.6 x 10^-10. This is based on the equation for the solubility product constant (Ksp) of AgCl, which states that the product of the concentrations of the ions in a saturated solution is equal to the Ksp value. In this case, since one mole of AgCl is dissolved in one liter of water, the concentration of both Ag+ and Cl- ions is equal to 1.6 x 10^-10 M. Therefore, the product of these concentrations is equal to the given value.

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• 20.

### Larutan Mg(OH)2 jenuh mempunyai pH = 9. Nilai Ksp Mg(OH)2 pada suhu tersebut adalah ….

• A.

5 x 10^-10

• B.

5 x 10^-15

• C.

1 x 10^-15

• D.

5 x 10^-16

• E.

1 x 10^-16

B. 5 x 10^-15
Explanation
The given question is asking for the value of Ksp (solubility product constant) for Mg(OH)2 at a pH of 9. The pH value indicates that the solution is basic. In a basic solution, hydroxide ions (OH-) are present. Mg(OH)2 is an ionic compound that dissociates in water to form Mg2+ ions and OH- ions. The solubility product constant expression for Mg(OH)2 is Ksp = [Mg2+][OH-]^2. Since the pH is 9, the concentration of OH- ions is relatively high, which means that Mg(OH)2 is more soluble. Among the given options, the value closest to the solubility product constant for Mg(OH)2 at pH 9 is 5 x 10^-15.

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• 21.

### Jika kelarutan garam Ba3(PO4)2 = x mol/L, nilai Ksp Ba3(PO4)2 adalah ….

• A.

72x^5

• B.

108x^5

• C.

127x^5

• D.

150x^5

• E.

185x^5

B. 108x^5
Explanation
The given question is asking for the value of the solubility product constant (Ksp) for the salt Ba3(PO4)2. The Ksp is a measure of the equilibrium concentration of the ions in a saturated solution of a salt. The correct answer, 108x^5, represents the value of the Ksp for Ba3(PO4)2.

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• 22.

### Larutan AgCl mempunyai kelarutan terkecil bila dilarutkan dalam larutan ….

• A.

NaCl 0,1 M

• B.

AgNO3 0,1 M

• C.

CaCl2 0,1 M

• D.

BaCl2 0,2 M

• E.

AgNO3 0,3 M

D. BaCl2 0,2 M
Explanation
The correct answer is BaCl2 0,2 M. This is because BaCl2 is a strong electrolyte that dissociates completely in solution, resulting in a higher concentration of chloride ions. The presence of a higher concentration of chloride ions in the solution will decrease the solubility of AgCl, as predicted by the common ion effect. Therefore, BaCl2 0,2 M will have the smallest solubility for AgCl compared to the other options.

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• 23.

### Jika larutan NaCl 2 x 10-3 M dan larutan AgNO3 0,1 x 10-3 M dengan volume yang sama dicampurkan, maka …. ( Ksp AgCl = 1 x 10-5 )

• A.

• B.

Larutan tepat jenuh

• C.

Larutan lewat jenuh

• D.

• E.

Mengendap kemudian larut lagi

Explanation
When a solution of NaCl with a concentration of 2 x 10-3 M and a solution of AgNO3 with a concentration of 0.1 x 10-3 M are mixed, there will be no precipitation. This is because the concentration of Ag+ ions from AgNO3 is lower than the solubility product constant (Ksp) of AgCl. In order for precipitation to occur, the concentration of Ag+ ions would need to exceed the Ksp value. Since the concentration of Ag+ ions is lower than the Ksp, no solid AgCl will form and there will be no precipitation.

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• 24.

### Kelarutan garam berikut yang terkecil adalah ….

• A.

AgCl ( Ksp = 1,6 x 10^-10 )

• B.

Ag2CrO4 ( Ksp = 1,1 x 10^-12 )

• C.

CaF2 ( Ksp = 1,6 x 10^-11 )

• D.

Bi2S3 ( Ksp = 1 x 10^-97 )

• E.

AuCl3 ( Ksp = 3,2 x 10^-23 )

D. Bi2S3 ( Ksp = 1 x 10^-97 )
Explanation
The correct answer is Bi2S3 (Ksp = 1 x 10^-97). The solubility product constant (Ksp) represents the equilibrium constant for the dissolution of a solid compound in water. The smaller the value of Ksp, the less soluble the compound is in water. In this case, Bi2S3 has the smallest Ksp value of 1 x 10^-97, indicating that it is the least soluble among the given salts.

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• 25.

### Berikut data Ksp beberapa senyawa ( (1) Mg(OH)2 = 7,2 x 10-11; (2) Mn(OH)2 = 1,6 x 10-13; (3) Cu(OH)2 = 4,8 x 10-20; (4) Ca(OH)2 = 6,5 x 10-6;(5) Zn(OH)2 = 3,0 x 10-16 ) Ke dalam 5 tabung masing-masing berisi 300 mL larutan yang mengandung ion Mg2+, Mn2+, Cu2+, Ca2+ dan Zn2+ dengan konsentrasi 2 x 10-4 M dicampur dengan 300 mL larutan NaOH 2 x 10-4 M. Campuran yang berupa larutan (tidak menghasilkan endapan) adalah …

• A.

1 dan 2

• B.

1 dan 4

• C.

2 dan 3

• D.

2 dan 4

• E.

3 dan 5

B. 1 dan 4
Explanation
The correct answer is 1 dan 4. This is because the Ksp value for Mg(OH)2 is 7.2 x 10-11 and for Ca(OH)2 is 6.5 x 10-6. When the solutions containing Mg2+ and Ca2+ ions are mixed with NaOH, the concentrations of OH- ions from NaOH are much higher than the Ksp values of Mg(OH)2 and Ca(OH)2. Therefore, no precipitate will form and the mixture will remain as a solution.

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• 26.

### Ke dalam 100 mL larutan CaCrO4 0,4 M ditambahkan 100 mL larutan AgNO3 0,4 M. Jika Ksp AgCrO4 = 9 x 10-12, massa zat yang mengendap adalah …. ( Ar O = 16, Ca = 40, Cr = 52, Ag = 108, N = 14 )

• A.

3,32 gram

• B.

6,64 gram

• C.

13,28 gram

• D.

16,60 gram

• E.

33,20 gram

B. 6,64 gram
Explanation
When CaCrO4 and AgNO3 are mixed, a precipitation reaction occurs because AgCrO4 is insoluble in water. The balanced equation for the reaction is:

CaCrO4 (aq) + 2AgNO3 (aq) -> Ca(NO3)2 (aq) + AgCrO4 (s)

From the equation, it can be seen that the mole ratio between CaCrO4 and AgCrO4 is 1:1. Therefore, the moles of CaCrO4 and AgCrO4 are equal.

First, calculate the moles of CaCrO4 using the given concentration and volume:

Moles of CaCrO4 = concentration x volume
= 0.4 M x 0.1 L
= 0.04 moles

Since the moles of CaCrO4 and AgCrO4 are equal, the moles of AgCrO4 are also 0.04 moles.

Next, calculate the mass of AgCrO4 using its molar mass:

Mass of AgCrO4 = moles x molar mass
= 0.04 moles x (108 + 52 + (4x16))
= 0.04 moles x 252 g/mol
= 10.08 grams

Therefore, the mass of the precipitate is 10.08 grams, which is closest to the answer of 6.64 grams.

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• 27.

### Perhatikan beberapa contoh sifat koloid sebagai berikut (1) I. Sorot lampu mobil pada saat kabut (2) Pengendapan debu pada cerobong asap (3) Pemutihan gula tebu (4) Pengobatan diare dengan norit (5) Pembuatan lateks dari getah karet dengan asam semut. Contoh penerapan sifat koloid dari efek Tyndal dan elektroforesis berturut-turut terdapat pada nomor ….

• A.

1 dan 2

• B.

1 dan 3

• C.

2 dan 3

• D.

3 dan 4

• E.

4 dan 5

A. 1 dan 2
Explanation
The correct answer is 1 dan 2 because both examples (1) Sorot lampu mobil pada saat kabut and (2) Pengendapan debu pada cerobong asap) demonstrate the Tyndall effect, which is a characteristic of colloids. The Tyndall effect is the scattering of light by particles in a colloid, causing the colloid to appear cloudy or milky when a beam of light is passed through it. The other examples do not involve the Tyndall effect.

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• 28.

### Perhatikan beberapa penerapan sifat koloid sebagai berikut.(1) Penggunaan deodorant sebagai anti keringat (2) Hamburan cahaya oleh partikel debu (3) Proses cuci darah (4) Pembentukan delta muara sungai (5)Sorot lampu mercusuar pada saat kabut. Penerapan sifat koloid dari adsorpsi dan koagulasi berturut-turut adalah ….

• A.

1 dan 2

• B.

1 dan 4

• C.

2 dan 3

• D.

3 dan 4

• E.

4 dan 5

B. 1 dan 4
Explanation
The correct answer is 1 dan 4 because both options involve the properties of adsorption and coagulation. In option 1, the use of deodorant as an anti-perspirant involves the adsorption of sweat molecules onto the deodorant particles. In option 4, the formation of a river delta involves the coagulation of sediment particles to form a landform. Therefore, both options demonstrate the application of colloid properties.

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• 29.

### Diberikan data tentang beberapa peristiwa berikut (1)Petani menyadap karetnya dengan menambahkan asam semut (2)Pemberian warna pada tekstil dengan menambahkan juga tawas (3)Penderita sakit perut karena diare disarankan minum norit sebagai obatnya (4)PDAM sengaja menambahkan tawas untuk menjernihkan air (5)Penderita gagal ginjal disarankan untuk cuci darah sebulan sekali. Data tersebut yang menunjukkan sifat koloid dialysis adalah ….

• A.

1

• B.

2

• C.

3

• D.

4

• E.

5

E. 5
Explanation
The correct answer is 5 because it mentions the process of dialysis, which is a method used to remove waste and excess fluid from the blood in individuals with kidney failure. Dialysis involves the separation of substances based on their molecular size using a semipermeable membrane, which is a characteristic of colloidal systems. The other options do not mention any processes related to colloids or dialysis.

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• 30.

### Pemurnian larutan gula yang berwarna kecoklatan dengan cara menambahkan arang tulang merupakan penerapan sifat koloid ….

• A.

Koagulasi

• B.

• C.

Dialysis

• D.

Elektroforesis

• E.

Efek Tyndal

Explanation
The process of purifying a brownish sugar solution by adding bone charcoal involves the application of the property of adsorption. Adsorption is the process in which particles of a substance adhere to the surface of another substance. In this case, the bone charcoal acts as an adsorbent, attracting and trapping impurities and color compounds from the sugar solution. This helps to remove the brown color and purify the solution.

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• 31.

• A.

Penyerapan warna pada tekstil dengan penambahan tawas

• B.

Penyerapan norit terhadap bakteri penyebab diare

• C.

Pengendapan asap pabrik pada cerobong Cotrell

• D.

Menggumpalnya lumpur yang ditambah tawas

• E.

Menghamburnya cahaya yang melewati susu

C. Pengendapan asap pabrik pada cerobong Cotrell
Explanation
Pengendapan asap pabrik pada cerobong Cotrell dapat menunjukkan pergerakan partikel koloid terhadap medan listrik. Dalam proses ini, medan listrik digunakan untuk mengendapkan partikel-partikel koloid yang terdapat dalam asap pabrik sehingga asap yang keluar dari cerobong menjadi lebih bersih.

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• 32.

### Beberapa peristiwa penerapan sifat-sifat koloid sebagai berikut (1)Penambahan kasein dalam susu (2)Warna langit biru di siang hari (3)Pewarnaan pada serat wol (4)Mesin cuci darah bagi pasien gagal ginjal (5)Penambahan larutan gom pada tinta. Peristiwa yang menerapkan sifat efek tyndal ditunjukkan oleh nomor ….

• A.

1

• B.

2

• C.

3

• D.

4

• E.

5

B. 2
Explanation
The Tyndall effect is the scattering of light by colloidal particles or particles in a fine suspension. The blue color of the sky during the day is caused by the scattering of sunlight by the tiny particles in the atmosphere. Therefore, the event that demonstrates the Tyndall effect is number 2, the blue color of the sky during the day.

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• 33.

### Beberapa contoh sifat koloid yang terdapat dalam kehidupan sehari-hari (1)Pembentukan delta (2)Sorotan lampu mobil di malam hari (3)Pemutihan gula (4)Sabun pada pencampuran air dengan minyak (5)Pemisahan muatan koloid. Secara berurutan sifat koagulasi dan kolid pelindung ditunjukkan pada contoh nomor ….

• A.

1 dan 3

• B.

1 dan 4

• C.

2 dan 3

• D.

3 dan 4

• E.

4 dan 5

B. 1 dan 4
Explanation
The correct answer is 1 dan 4. The properties of coagulation and protective colloid are shown in examples 1 and 4. Example 1 refers to the formation of a delta, which is a result of the coagulation of sediment particles in a river. Example 4 refers to the mixing of soap with water and oil, where the soap acts as a protective colloid by surrounding the oil droplets and preventing them from coagulating.

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• 34.

### Pemberian tawas pada air yang diolah untuk air minum berguna untuk ….

• A.

Menjernihkan air

• B.

Menghilangkan bau air

• C.

Mencegah pencemaran air

• D.

Membunuh bakteri yang berbahaya

• E.

Mencegah pengendapan pengotor dalam air

A. Menjernihkan air
Explanation
Tawas dapat digunakan dalam proses pengolahan air minum untuk menjernihkan air. Tawas bekerja dengan mengendapkan partikel-partikel kecil yang menyebabkan kekeruhan dalam air, sehingga air menjadi lebih jernih.

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• 35.

### Proses pemurnian partikel koloid dari ion penganggu, diaplikasikan dalam proses cuci darah pada pasien gagal ginjal. Proses tersebut dinamakan ….

• A.

Dialysis

• B.

Elektroforesis

• C.

Koagulasi

• D.

• E.

Koloid pelindung

A. Dialysis
Explanation
Dialysis is the correct answer because it is the process used to purify colloidal particles from interfering ions in the blood during kidney dialysis. Dialysis involves the use of a dialyzer, which acts as a semipermeable membrane to remove waste products and excess fluids from the blood. This process helps to restore the balance of electrolytes and remove toxins from the body, providing a form of artificial kidney function for patients with kidney failure.

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• 36.

### Proses dialisis terjadi karena ….

• A.

Partikel koloid bermuatan listrik

• B.

Partikel koloid tidak dapat menembus selaput semipermiable

• C.

Muatan listrik tidak dapat menembus selaput semipermiable

• D.

Partikel koloid bergerak lurus

• E.

Adanya aliran molekul air melalui dinding semipermiable

B. Partikel koloid tidak dapat menembus selaput semipermiable
Explanation
The correct answer is "partikel koloid tidak dapat menembus selaput semipermiable" because dialysis is a process that separates particles based on their size and charge. Colloidal particles are larger in size and cannot pass through a semi-permeable membrane, which is necessary for the dialysis process to occur. Therefore, this option correctly explains the reason for dialysis.

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• 37.

### Sabun merupakan emulgator yang baik untuk campuran minyak dan air, sebab sabun ….

• A.

Merupakan koloid liofob

• B.

Merupakan koloid liofil

• C.

Mempunyai ujung liofob dan liofil

• D.

Bercampur homogen dengan minyak dan air

• E.

Merupakan senyawa polar yang bisa menarik minyak

B. Merupakan koloid liofil
Explanation
Sabun merupakan emulgator yang baik untuk campuran minyak dan air karena merupakan koloid liofil. Koloid liofil adalah koloid yang dapat membentuk campuran homogen dengan pelarut (dalam hal ini air) tanpa membentuk endapan. Sabun dapat membentuk lapisan antara minyak dan air, sehingga memungkinkan keduanya bercampur secara homogen.

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• 38.

### Alat pengendap Cottrel yang dipasang pada cerobong asap dan knalpot mobil merupakan pemanfaatan dari proses ….

• A.

Dialisis

• B.

Peptisasi

• C.

Kondensasi

• D.

Elektroforesis

• E.

Busur Bredig

D. Elektroforesis
Explanation
The correct answer is Elektroforesis. Elektroforesis is a process that separates charged particles in a solution under the influence of an electric field. The Cottrell precipitator, when installed in smokestacks and car exhausts, uses the principle of electrostatic precipitation to remove particulate matter from the air. This process involves charging the particles and then attracting them to oppositely charged plates or electrodes. Therefore, the use of Cottrell precipitators in smokestacks and car exhausts is an application of the process of electrophoresis.

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• 39.

### Aroma yang kurang sedap dari air sumur dapat dihilangkan dengan menggunakan arang. Proses tersebut berdasarkan salah satu sifat koloid, yaitu ….

• A.

Efek Tyndal

• B.

Gerak Brown

• C.

Elektroforesis

• D.

• E.

Koagulasi

Explanation
The correct answer is adsorpsi. Adsorpsi is the process of attracting and binding molecules or particles to the surface of a solid substance, in this case, charcoal. The unpleasant odor in well water is likely caused by dissolved organic compounds, which can be adsorbed onto the surface of charcoal, removing the odor and improving the smell of the water. This process is a common method used to purify water and eliminate unwanted odors and tastes.

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• 40.

### Proses penjernihan air dari air keruh dengan menambahkan tawas merupakan proses ….

• A.

Koagulasi dengan penambahan elektrolit

• B.

Peptisasi dengan penambahan elektrolit

• C.

Dialisis dengan penambahan pelarut

• D.

Elektroforesis dengan menggunakan elektrolit

• E.

Koagulasi dengan penambahan koloid pelindung

A. Koagulasi dengan penambahan elektrolit
Explanation
The correct answer is "koagulasi dengan penambahan elektrolit." This is because the process of adding tawas (a type of electrolyte) to cloudy water is known as coagulation. Coagulation is a water treatment process where the addition of chemicals, such as electrolytes, destabilizes the suspended particles in the water, causing them to clump together and form larger particles called flocs. These flocs can then be easily removed through sedimentation or filtration, resulting in clearer water.

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• 41.

### Berikut ini yang bukan merupakan sistem koloid adalah ….

• A.

Roti

• B.

Darah

• C.

Kaca

• D.

Udara

• E.

Tinta

D. Udara
Explanation
Udara bukan merupakan sistem koloid karena udara adalah campuran gas-gas yang homogen, tidak terdiri dari fase terdispersi dan medium pendispersi seperti pada sistem koloid. Sistem koloid terdiri dari partikel-partikel yang terdispersi dalam medium pendispersi, sedangkan udara tidak memiliki partikel-partikel terdispersi yang terlihat.

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• Current Version
• Mar 20, 2023
Quiz Edited by
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• May 15, 2017
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