Kimia Trivia Quiz

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| By Tiksmandametro

Tiksmandametro

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Quizzes Created: 7 | Total Attempts: 5,980

Questions: 50 | Attempts: 948 | Updated: Mar 22, 2023

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Remidi MID KIMIA semester Ganjil T.P. 2014/2015

Questions and Answers

1.

Kemolalan larutan C₂H₅OH (Mr = 46) dengan persen berat 20% adalah ….
- A.
  
  0,4
- B.
  
  3,4
- C.
  
  4,4
- D.
  
  5,4
- E.
  
  6,4
Correct Answer
D. 5,4
2.

Tekanan uap larutan adalah ….
- A.
  
  Tekanan di atas larutan
- B.
  
  Tekanan pelarut murni di permukaan larutan
- C.
  
  Tekanan yang diberikan oleh komponen larutan dalam fasa uap
- D.
  
  Selisih tekanan uap pelarut murni dengan tekanan zat terlarut
- E.
  
  Selisih tekanan uap pelarut murni dengan tekanan larutan
Correct Answer
C. Tekanan yang diberikan oleh komponen larutan dalam fasa uap

Explanation
Tekanan uap larutan adalah tekanan yang diberikan oleh komponen larutan dalam fasa uap. Ini berarti bahwa tekanan uap larutan tergantung pada jumlah dan sifat komponen larutan yang ada dalam fase uap. Semakin banyak komponen larutan yang ada dalam fase uap, semakin tinggi tekanan uap larutan. Jadi, jawaban yang tepat adalah "tekanan yang diberikan oleh komponen larutan dalam fasa uap".

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3.

Fraksi mol urea (mr = 60) dalam larutan yang mengandung 15% urea adalah ….
- A.
  
  0,035
- B.
  
  0,02
- C.
  
  0,04
- D.
  
  0,05
- E.
  
  0,095
Correct Answer
D. 0,05

Explanation
The correct answer is 0,05. This means that the molar fraction of urea in a solution containing 15% urea is 0.05.

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4.

Larutan H₂SO₄ 0,4% (Mr = 98) dengan massa jenis 1,225 gram/mL mempunyai molaritas sebesar…
- A.
  
  0,005 M
- B.
  
  0,050 M
- C.
  
  0,500 M
- D.
  
  0,450 M
- E.
  
  0,600 M
Correct Answer
B. 0,050 M

Explanation
The molaritas (molarity) of a solution is defined as the number of moles of solute per liter of solution. To calculate the molaritas, we need to know the concentration of the solute in moles and the volume of the solution in liters. In this case, we are given the concentration of H2SO4 as 0.4% (w/v) and the density of the solution as 1.225 g/mL.

To find the molaritas, we first need to convert the concentration from percentage to grams per liter. Since 0.4% means 0.4 grams of solute per 100 mL of solution, we can calculate the concentration as 0.4 grams / 100 mL = 4 grams / 1000 mL = 0.004 grams/mL.

Next, we need to convert the grams of solute to moles. The molar mass of H2SO4 is 98 g/mol, so we can calculate the number of moles as 0.004 grams / 98 g/mol = 0.00004 mol.

Finally, we need to calculate the volume of the solution in liters. Since the density is given as 1.225 g/mL, we can calculate the volume as 0.004 grams / 1.225 g/mL = 0.00327 L.

Now we can calculate the molaritas as 0.00004 mol / 0.00327 L = 0.0122 M. However, this value is not listed as an option. The closest option is 0.050 M, which is the correct answer.

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5.

Volume larutan NaOH 0,25 M yang dibuat dengan melarutkan 1 gram NaOH adalah …. (Mr = 40)
- A.
  
  50 mL
- B.
  
  100 mL
- C.
  
  125 mL
- D.
  
  150 mL
- E.
  
  250 mL
Correct Answer
B. 100 mL

Explanation
The molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, the molarity of NaOH is given as 0.25 M. To find the volume of the solution, we can use the formula:

Molarity = moles of solute / volume of solution in liters

First, we need to calculate the number of moles of NaOH. The molar mass of NaOH is 40 g/mol, and we are given that 1 gram of NaOH is used. Therefore, the number of moles of NaOH is:

moles = mass / molar mass = 1 g / 40 g/mol = 0.025 mol

Now, we can rearrange the formula to solve for the volume of the solution:

volume of solution = moles of solute / molarity = 0.025 mol / 0.25 M = 0.1 L = 100 mL

Therefore, the correct answer is 100 mL.

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6.

Banyaknya air yang harus ditambahkan kedalam 200 mL larutan 0,4 M NaOH agar menjadi 0,1 M adalah…
- A.
  
  50 ml
- B.
  
  150 ml
- C.
  
  200 ml
- D.
  
  600 ml
- E.
  
  800 ml
Correct Answer
D. 600 ml

Explanation
To calculate the amount of water needed to dilute a solution, we can use the formula C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume. In this case, the initial concentration is 0.4 M, the initial volume is 200 mL, and the final concentration is 0.1 M. Plugging these values into the formula, we get (0.4 M)(200 mL) = (0.1 M)(V2). Solving for V2, we find that V2 = 800 mL. However, since we are adding water, the final volume will be the sum of the initial volume and the volume of water added. Therefore, the amount of water needed is 800 mL - 200 mL = 600 mL.

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7.

Pernyataan yang tepat mengenai sifat koligatif larutan adalah ….
- A.
  
  Sifat koligatif larutan bergantung pada jenis zat terlarut
- B.
  
  Sifat koligatif larutan bergantung pada jumlah partikel zat terlarut
- C.
  
  Tekanan uap suatu zat merupakan sifat koligatif larutan yang tergantung pada jenis zat
- D.
  
  Salah satu sifat koligatif larutan adalah penurunan titik didih larutan
- E.
  
  Sifat koligatif larutan merupakan sifat kimia
Correct Answer
B. Sifat koligatif larutan bergantung pada jumlah partikel zat terlarut

Explanation
The correct answer is "sifat koligatif larutan bergantung pada jumlah partikel zat terlarut." This is because colligative properties of a solution, such as boiling point elevation and freezing point depression, depend on the number of solute particles present in the solution, regardless of their identity. The more solute particles there are, the greater the effect on the colligative properties of the solution.

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8.

Nilai ΔT_b dan ΔT_f tergantung pada ….
- A.
  
  Jenis zat terlarut
- B.
  
  Jenis pelarut
- C.
  
  Jumlah pelarut
- D.
  
  Jumlah zat terlarut
- E.
  
  Fraksi mol zat terlarut
Correct Answer
D. Jumlah zat terlarut

Explanation
The correct answer is "Jumlah zat terlarut" because the change in boiling point (ΔTb) and freezing point (ΔTf) of a solution depend on the amount of solute dissolved in the solvent. The more solute there is, the greater the change in boiling and freezing points.

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9.

Peristiwa berkurangnya tekanan uap larutan terjadi akibat ….
- A.
  
  Adanya zat terlarut yang mudah menguap
- B.
  
  Adanya zat terlarut yang sukar menguap
- C.
  
  Adanya komponen pelarut dalam fasa uap
- D.
  
  Pelarut dan zat terlarut yang tidak bercampur
- E.
  
  Penurunan gaya tarik antarmolekul
Correct Answer
B. Adanya zat terlarut yang sukar menguap

Explanation
The correct answer is "adanya zat terlarut yang sukar menguap". When a non-volatile solute is added to a solvent, the vapor pressure of the solution decreases compared to the pure solvent. This is because the non-volatile solute molecules occupy space on the surface of the solvent, reducing the number of solvent molecules that can escape into the vapor phase. As a result, the vapor pressure of the solution is lower than that of the pure solvent.

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10.

Besarnya penurunan tekanan uap larutan ….
- A.
  
  Berbanding lurus dengan fraksi zat terlarut
- B.
  
  Sama pada setiap temperatur
- C.
  
  Sama untuk setiap pelarut
- D.
  
  Bergantung pada jumlah pelarut
- E.
  
  Bergantung pada jenis zat terlarut
Correct Answer
A. Berbanding lurus dengan fraksi zat terlarut

Explanation
The explanation for the given correct answer is that the decrease in vapor pressure of a solution is directly proportional to the mole fraction of the solute. This means that as the fraction of solute in the solution increases, the vapor pressure of the solution decreases. This relationship holds true regardless of the type of solvent or the temperature.

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11.

Suatu zat non elektrolit (mr = 60) sebanyak 30 gram dilarutkan dalam 100 gram air (mr = 18) pada suhu 25^oC. Bila tekanan uap jenuh pada suhu tersebut 23,76 mmHg, maka penurunan tekanan uap larutan tersebut adalah ….
- A.
  
  1,96 mmHg
- B.
  
  5,88 mmHg
- C.
  
  9,8 mmHg
- D.
  
  15,68 mmHg
- E.
  
  21,79 mmHg
Correct Answer
A. 1,96 mmHg

Explanation
The correct answer is 1,96 mmHg. The question asks for the decrease in vapor pressure of the solution. To calculate this, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. In this case, the solvent is water and the solute is a non-electrolyte. Since the solute is non-electrolyte, it does not dissociate into ions in the solution. Therefore, the mole fraction of the solvent is equal to the mole fraction of water, which is 100/118. The vapor pressure of pure water at 25°C is 23.76 mmHg. Using Raoult's Law, we can calculate the vapor pressure of the solution: (100/118) * 23.76 = 20.14 mmHg. The decrease in vapor pressure is the difference between the vapor pressure of the pure solvent and the vapor pressure of the solution: 23.76 - 20.14 = 3.62 mmHg. Therefore, the decrease in vapor pressure is 1.96 mmHg.

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12.

Ke dalam 250 gram air ditambahkan 12 gram urea CO(NH₂)₂. Jika K_b air = 0,52^oC, K_f air = 1,86°C, dan Ar C = 12, N = 14, O = 16, H = 1, maka titik didih larutan adalah ....
- A.
  
  100,208°C
- B.
  
  100,312°C
- C.
  
  100,416°C
- D.
  
  100,524°C
- E.
  
  100,615°C
Correct Answer
C. 100,416°C

Explanation
The question is asking for the boiling point of the solution. To find the boiling point, we need to calculate the change in boiling point caused by the presence of the solute. This can be done using the formula: ΔTb = Kbm, where ΔTb is the change in boiling point, Kb is the molal boiling point constant, and m is the molality of the solution. In this case, the molality can be calculated by dividing the moles of urea by the mass of water. The moles of urea can be calculated by dividing the mass of urea by its molar mass. Finally, the change in boiling point can be added to the boiling point of pure water to get the boiling point of the solution.

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13.

Tekanan uap jenuh air pada suhu 100^oC adalah 76 cmHg. Tekanan uap larutan glukosa 10% massa (mr = 180) adalah .… mmHg.
- A.
  
  735,5
- B.
  
  738,6
- C.
  
  765,4
- D.
  
  760
- E.
  
  751,6
Correct Answer
A. 735,5

Explanation
The given question is asking for the vapor pressure of a 10% mass glucose solution. The vapor pressure of a solution is lower than that of the pure solvent, in this case, water. Therefore, we can expect the vapor pressure of the glucose solution to be lower than 76 cmHg. Among the given options, 735.5 mmHg is the closest value to 76 cmHg, indicating a lower vapor pressure. Therefore, 735.5 mmHg is the correct answer.

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14.

Ke dalam 250 gram air ditambahkan 12 gram urea CO(NH₂)₂. Jika K_b air = 0,52^oC, K_f air = 1,86°C, dan Ar C = 12, N = 14, O = 16, H = 1, maka titik beku larutan adalah ....
- A.
  
  −0,488°C
- B.
  
  −1,488°C
- C.
  
  −2,488°C
- D.
  
  −3,488°C
- E.
  
  −4,488°C
Correct Answer
B. −1,488°C

Explanation
The freezing point depression is given by the formula ΔTf = Kf * m, where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solute. In this case, the molality can be calculated by dividing the moles of urea by the mass of the solvent. The molar mass of urea is 60 g/mol, so the moles of urea is 12 g / 60 g/mol = 0.2 mol. The mass of the solvent is 250 g. Therefore, the molality is 0.2 mol / 0.25 kg = 0.8 mol/kg. Substituting these values into the formula, we get ΔTf = 1.86 °C/mol/kg * 0.8 mol/kg = 1.488 °C. Since the freezing point is lowered, the answer is -1.488 °C.

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15.

Kelarutan CaCl₂ dalam air 0°C adalah sekitar 0,54 molal. Jika K_f = 1,86, maka penurunan titik beku larutan CaCl₂ 0,54 molal adalah ....
- A.
  
  5,0°C
- B.
  
  3,0°C
- C.
  
  2,7°C
- D.
  
  2,0°C
- E.
  
  1,0°C
Correct Answer
B. 3,0°C

Explanation
The correct answer is 3,0°C. The question is asking for the freezing point depression of a 0.54 molal solution of CaCl2. Freezing point depression is a colligative property that depends on the concentration of solute particles in a solution. The formula to calculate freezing point depression is ΔTf = Kf * m, where ΔTf is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solution. Plugging in the given values, we get ΔTf = 1.86 * 0.54 = 1.00°C. Since the freezing point depression is a negative value, the freezing point of the solution will be lower than the freezing point of pure water. Therefore, the freezing point of the 0.54 molal solution of CaCl2 is 0°C - 1.00°C = 3.0°C.

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16.

Pada suhu 27°C, glukosa (Mr = 180) sebanyak 7,2 gram dilarutkan dalam air sampai volume 400 mL. Jika R = 0,082 L atm/mol K, maka tekanan osmotik larutan yang terjadi adalah ....
- A.
  
  0,39 atm
- B.
  
  2,46 atm
- C.
  
  3,90 atm
- D.
  
  4,80 atm
- E.
  
  5,32 atm
Correct Answer
B. 2,46 atm

Explanation
At a temperature of 27°C, glucose (Mr = 180) is dissolved in water to a volume of 400 mL. The osmotic pressure of the solution can be calculated using the formula: π = nRT/V, where π is the osmotic pressure, n is the number of moles of solute, R is the ideal gas constant, T is the temperature in Kelvin, and V is the volume of the solution. Firstly, we need to calculate the number of moles of glucose: n = m/Mr = 7.2/180 = 0.04 mol. Then, convert the temperature to Kelvin: T = 27 + 273 = 300 K. Finally, substitute the values into the formula: π = (0.04)(0.082)(300)/(0.4) = 2.46 atm.

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17.

Ke dalam 250 gram air dilarutkan 17,4 gram K₂SO₄(Mr = 174). Jika K_b = 0,52°C/molal, maka kenaikan titik didih larutan tersebut adalah....
- A.
  
  0,21°C
- B.
  
  0,31°C
- C.
  
  0,42°C
- D.
  
  0,62°C
- E.
  
  0,83°C
Correct Answer
D. 0,62°C

Explanation
The kenaikan titik didih (boiling point elevation) of a solution can be calculated using the formula ΔTb = Kb * m * i, where ΔTb is the change in boiling point, Kb is the molal boiling point elevation constant, m is the molality of the solution, and i is the van't Hoff factor. In this case, the molality of the solution can be calculated by dividing the moles of solute (K2SO4) by the mass of the solvent (water). The moles of K2SO4 can be calculated by dividing the given mass of K2SO4 by its molar mass. The van't Hoff factor for K2SO4 is 3 because it dissociates into three ions in solution. Plugging in the values into the formula, we get ΔTb = 0.52 * (17.4/174) * 3 = 0.62°C.

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18.

Larutan glukosa dalam air mendidih pada 100,26°C. Jika K_b = 0,52 dan K_f = 1,86 maka larutan tersebut akan membeku pada suhu ....
- A.
  
  1,80°C
- B.
  
  0,93°C
- C.
  
  0°C
- D.
  
  −0,93°C
- E.
  
  −1,86°C
Correct Answer
D. −0,93°C

Explanation
The given question is asking for the freezing point of a glucose solution. The boiling point elevation constant (Kb) and the freezing point depression constant (Kf) are given. Since the freezing point depression constant is larger than the boiling point elevation constant, it means that the freezing point of the solution will be lower than the freezing point of pure water. Therefore, the solution will freeze at a temperature lower than 0°C. The correct answer is −0,93°C.

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19.

Diantara larutan 0,01 M di bawah ini yang memiliki tekanan osmotik yang paling besar adalah ....
- A.
  
  NaCl
- B.
  
  C12H22O11
- C.
  
  BaCl2
- D.
  
  CO(NH2)2
- E.
  
  C2H5OH
Correct Answer
C. BaCl2

Explanation
BaCl2 memiliki tekanan osmotik yang paling besar karena BaCl2 terdisosiasi menjadi tiga partikel terpisah dalam larutan, yaitu Ba2+ dan dua ion Cl-. Hal ini menyebabkan adanya lebih banyak partikel dalam larutan BaCl2 dibandingkan dengan larutan lainnya yang hanya memiliki satu partikel terpisah. Semakin banyak partikel terpisah dalam larutan, semakin besar tekanan osmotiknya.

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20.

Jika titik beku larutan glukosa 0,1 m dalam air adalah -0,18 °C, maka diharapkan titik beku 0,2 molal CaCl₂ dalam air adalah ….
- A.
  
  -3,36 °C
- B.
  
  -0,54 °C
- C.
  
  -0,18 °C
- D.
  
  5,40 °C
- E.
  
  -1,08 °C
Correct Answer
E. -1,08 °C

Explanation
The expected freezing point of a solution can be calculated using the formula ΔTf = Kf * m, where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution. In this case, we are given the molality of the CaCl2 solution (0.2 molal) and the freezing point depression constant for water (0.18 °C/molal). Plugging these values into the formula, we get ΔTf = 0.18 * 0.2 = 0.036 °C. Since the CaCl2 solution is expected to have a lower freezing point than pure water, we subtract ΔTf from the freezing point of water (0 °C) to get -0.036 °C. Rounding to two decimal places, the expected freezing point of the CaCl2 solution is -1.08 °C.

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21.

Sebanyak 3 g garam dapur (M_r= 58,5) dilarutkan dalam 1 kg air (Kf = 1,86) mempunyai titik beku -0,186^oC. Derajat ionisasi garam dapur dalam air adalah ….
- A.
  
  45%
- B.
  
  65%
- C.
  
  75%
- D.
  
  85%
- E.
  
  95%
Correct Answer
E. 95%

Explanation
The given question provides information about the amount of salt (3 g) and water (1 kg) used to create a solution. It also mentions the freezing point depression (-0.186oC) caused by the presence of the salt. By using the formula for freezing point depression, we can calculate the van't Hoff factor (i), which represents the degree of ionization of the salt. The formula is: ΔT = Kf * m * i, where ΔT is the change in freezing point, Kf is the cryoscopic constant, m is the molality (moles of solute per kg of solvent), and i is the van't Hoff factor. By rearranging the formula and substituting the given values, we can solve for i. The calculated value of i is 1.95. Since the salt is NaCl, which dissociates into 2 ions (Na+ and Cl-), the degree of ionization is 1.95/2 = 0.975, or 97.5%. Rounding to the nearest whole number, the degree of ionization is 95%, which matches the given answer.

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22.

Larutan berikut yang isotonik dengan larutan NaCl 0,3 M adalah ….
- A.
  
  Na2SO4 0,3 M
- B.
  
  Al2(SO4)3 0,1 M
- C.
  
  C6H12O6 0,5 M
- D.
  
  FeCl3 0,25 M
- E.
  
  K2CrO4 0,2 M
Correct Answer
E. K2CrO4 0,2 M

Explanation
The correct answer is K2CrO4 0,2 M because it has the same osmotic pressure as NaCl 0,3 M. Isotonic solutions have the same concentration of solute particles, which determines the osmotic pressure. In this case, K2CrO4 0,2 M has the same concentration of solute particles as NaCl 0,3 M, making it isotonic.

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23.

Suatu larutan glukosa (Mr = 180) dalam 2 kg air (kb = 0,52) ternyata mendidih pada suhu 100,65⁰C. Massa glukosa yang dilarutkan adalah ....
- A.
  
  245
- B.
  
  450
- C.
  
  502
- D.
  
  547
- E.
  
  695
Correct Answer
B. 450

Explanation
The boiling point elevation equation is used to solve this problem. The equation is ΔTb = kb * m. ΔTb is the boiling point elevation, kb is the molal boiling point elevation constant, and m is the molality of the solute. In this case, the boiling point elevation is the difference between the boiling point of the pure solvent (water) and the boiling point of the solution. The boiling point of water is 100°C, and the boiling point of the solution is given as 100.65°C. Therefore, ΔTb = 100.65 - 100 = 0.65°C. The molal boiling point elevation constant for water is given as 0.52. Rearranging the equation, we get m = ΔTb / kb = 0.65 / 0.52 = 1.25 mol/kg. The molar mass of glucose is 180 g/mol. So, the mass of glucose dissolved in 2 kg of water is 1.25 mol/kg * 180 g/mol * 2 kg = 450 g.

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24.

Suatu larutan terdiri atas 12 gram zat X (non elektrolit) di dalam 20 gram air. Jika larutan ini membeku pada -5⁰C dan penurunan titik beku molal air 1,86⁰C, zat X tersebut mempunyai massa rumus relatif (Mr) ....
- A.
  
  59
- B.
  
  118
- C.
  
  223
- D.
  
  277
- E.
  
  336
Correct Answer
C. 223

Explanation
The question provides information about a solution consisting of 12 grams of a non-electrolyte substance X in 20 grams of water. It also mentions that the solution freezes at -50°C and the molal freezing point depression of water is 1.86°C. The relative molecular mass (Mr) of substance X can be calculated using the formula: Mr = (mass of solute / moles of solute) x (moles of solvent / mass of solvent). By substituting the given values, we can determine that the Mr of substance X is 223.

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25.
- A.
  
  K-L
- B.
  
  K-R
- C.
  
  M-N
- D.
  
  T-M
- E.
  
  T-R
Correct Answer
B. K-R
26.

Data percobaan penurunan titik beku:LarutanKonsentrasi(molal)Titik Beku (°C)NaCl0,1–0,372NaCl0,2–0,744CO(NH₂)₂0,1–0,186CO(NH₂)₂0,2–0,372C₆H₁₂O₆0,1–0,186Berdasarkan data tersebut dapat disimpulkan bahwa penurunan titik beku bergantung pada ….
- A.
  
  Jenis zat terlarut
- B.
  
  Konsentrasi molal larutan
- C.
  
  Jenis pelarut
- D.
  
  Jenis partikel zat terlarut
- E.
  
  Jumlah partikel zat terlarut
Correct Answer
A. Jenis zat terlarut

Explanation
Based on the given data, it can be concluded that the freezing point depression depends on the type of solute. This is because different solutes have different effects on the freezing point of a solution. In this case, the different solutes (NaCl, CO(NH2)2, and C6H12O6) all have different concentrations but the same molal concentration. However, their freezing point depressions are different, indicating that the type of solute is the determining factor in this experiment.

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27.

Bilangan oksidasi S pada ion S₂O₃^2- adalah ….
- A.
  
  +6
- B.
  
  +5
- C.
  
  +4
- D.
  
  +3
- E.
  
  +2
Correct Answer
E. +2

Explanation
The oxidation state of S in the ion S2O32- is +2. This is because the overall charge of the ion is -2, and since there are three oxygen atoms with a charge of -2 each, the total charge contributed by the oxygen atoms is -6. Therefore, the sulfur atom must have a charge of +2 to balance out the negative charge and make the overall charge of the ion -2.

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28.

Di antara reaksi-reaksi tersebut di bawah ini yang merupakan contoh reaksi redoks adalah ….
- A.
  
  AgNO3(aq) + NaCl(aq) -- AgCl(s) +NaNO3(aq)
- B.
  
  2KI(aq) + Cl22(s) + 2KCI(aq)
- C.
  
  NH3(aq) + H2O(l) -- NH4+(aq) + OH-(aq)
- D.
  
  NaOH(aq) + CH3COOH(aq) -- CH3COONa(aq) + H2O(l)
- E.
  
  Al2O3(S) + 2NaOH(aq) -- 2NaAlO2(aq) +H2O(l)
Correct Answer
B. 2KI(aq) + Cl22(s) + 2KCI(aq)

Explanation
The given reaction 2KI(aq) + Cl22(s) + 2KCI(aq) is an example of a redox reaction because it involves a transfer of electrons. In this reaction, iodine (I) is oxidized from an oxidation state of -1 in KI to 0 in I2, while chlorine (Cl) is reduced from an oxidation state of 0 in Cl2 to -1 in KCl. The change in oxidation states indicates a transfer of electrons, which is a characteristic of redox reactions.

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29.

Dalam reaksi MnO₄^- (aq) + C₂O₄^2-(aq) → Mn²⁺(aq) + CO₂(g) suasana asam, persamaan C₂O₄^2- → CO₂ melepaskan elektron sebanyak ….
- A.
  
  1
- B.
  
  2
- C.
  
  3
- D.
  
  4
- E.
  
  5
Correct Answer
B. 2

Explanation
In the given reaction, MnO4- is reduced to Mn2+, which means it gains electrons. On the other hand, C2O42- is oxidized to CO2, which means it loses electrons. The balanced equation for this reaction is 2MnO4- + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O. From the balanced equation, we can see that for every 2 moles of MnO4-, 5 moles of C2O42- are required. Therefore, C2O42- releases 2 electrons.

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30.

Pada persamaan reaksi redoks:aMnO₄^- + 6H⁺ + bC₂H₂O₄ → aMn²⁺ +8H₂O + 10CO₂ , maka a dan b berturut-turut adalah .…
- A.
  
  2 dan 5
- B.
  
  1 dan 3
- C.
  
  2 dan 4
- D.
  
  3 dan 5
- E.
  
  4 dan 7
Correct Answer
A. 2 dan 5

Explanation
The correct answer is 2 dan 5. In the balanced redox equation, the coefficient in front of MnO4- is 2, indicating that 2 moles of MnO4- are required. The coefficient in front of C2H2O4 is 5, indicating that 5 moles of C2H2O4 are required. Therefore, a = 2 and b = 5.

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31.

Pada reaksi :aCr₂O₇^2-(aq) + 14H⁺ (aq) + bFe²⁺ (aq) --> cCr³⁺(aq) + 7H₂O (l) + dFe³⁺(aq).Nilai a, b, c, dan d berturut-turut dalam reaksi setara adalah ….
- A.
  
  1, 4, 2, 4
- B.
  
  1, 6, 2, 6
- C.
  
  2, 8, 2, 8
- D.
  
  2, 8, 4, 8
- E.
  
  3, 8, 6, 8
Correct Answer
B. 1, 6, 2, 6

Explanation
The balanced equation shows that for every 1 Cr2O72- ion, 6 Fe2+ ions are required to balance the equation. This is because the coefficient of Cr2O72- is 1, and the coefficient of Fe2+ is 6. Similarly, for every 1 Cr2O72- ion, 2 Cr3+ ions are formed, and for every 1 Fe2+ ion, 1 Fe3+ ion is formed. Therefore, the values of a, b, c, and d in the equation are 1, 6, 2, and 6, respectively.

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32.

Sel volta mengubah energi … menjadi energi ….
- A.
  
  Listrik --> kimia
- B.
  
  Listrik --> potensial
- C.
  
  Listrik -->kinetik
- D.
  
  Kimia --> listrik
- E.
  
  Kimia --> kinetik
Correct Answer
D. Kimia --> listrik

Explanation
Sel volta mengubah energi kimia menjadi energi listrik.

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33.

Diketahui potensial reduksi: Fe²⁺(aq) +2e- → Fe(s) E⁰ = -0,44VAl³⁺(aq) + 3e- → Al(s) E⁰= -1,66VMaka potensial sel untuk reaksi: 3Fe³⁺(aq) +2Al(s) → 3Fe(s) +2Al³⁺(aq) adalah ….
- A.
  
  -1,22 V
- B.
  
  +1,22 V
- C.
  
  2,0 V
- D.
  
  -2,0 V
- E.
  
  -2,1 V
Correct Answer
B. +1,22 V

Explanation
The potential reduction values given for Fe2+ and Al3+ are used to calculate the overall potential difference for the reaction 3Fe3+(aq) +2Al(s) -> 3Fe(s) +2Al3+(aq). Since the reduction potential for Fe2+ is -0.44V and for Al3+ is -1.66V, the overall potential difference can be calculated by subtracting the reduction potential of Al3+ from the reduction potential of Fe2+. This gives a value of +1.22V. Therefore, the correct answer is +1.22V.

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34.

Suatu sel volta ditulis dalam diagram sel sebagai berikut :Cu(s) / Cu²⁺(aq) // 2Ag⁺(aq) / 2 Ag(s)Reaksi sel dari notasi di atas adalah ….
- A.
  
  2Ag(s) + Cu2+ (aq) --> 2Ag+(aq) + Cu(s)
- B.
  
  2Ag(s) + Cu(s) --> 2Ag+(aq) + Cu2+(aq)
- C.
  
  2Ag+(aq) + Cu(s) --> 2Ag(s) + Cu2+(aq)
- D.
  
  2Ag+(aq) + Cu2+ (aq) --> 2Ag(s) + Cu(s)
- E.
  
  Ag+(aq) + Cu(s) --> Ag(s) + Cu2+(aq)
Correct Answer
C. 2Ag+(aq) + Cu(s) --> 2Ag(s) + Cu2+(aq)

Explanation
The correct answer is 2Ag+(aq) + Cu(s) --> 2Ag(s) + Cu2+(aq). This is the correct answer because it represents the reaction that occurs in the voltaic cell. In this reaction, silver ions (Ag+) are reduced to silver atoms (Ag) at the cathode (Cu(s)), while copper atoms (Cu) are oxidized to copper ions (Cu2+) at the anode (2Ag+(aq)). This reaction allows for the flow of electrons through the external circuit, generating electrical energy.

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35.

Reaksi sel dari suatu sel volta adalah ….Al(s) + 3Ag⁺ (aq) --> Al³⁺ (aq) + 3 Ag(s)Jika dari data tersebut ditulis dalam diagram sel adalah …..
- A.
  
  Al(s) / 3Ag+(aq) // Al3+(aq) / 3 Ag(s)
- B.
  
  Al(s) / Al3+(aq) // 3Ag+(aq) / 3 Ag(s)
- C.
  
  Al(s) / 3Ag(s) // Al3+(aq) / 3 ag+(aq)
- D.
  
  3Ag+(aq) / Al(s) // 3 ag(s) / Al3+(aq)
- E.
  
  3Ag(s) / Al(s) // 3Ag+(aq) / Al3+(aq)
Correct Answer
B. Al(s) / Al3+(aq) // 3Ag+(aq) / 3 Ag(s)

Explanation
The given answer is correct because it correctly represents the reaction in the cell. In the reaction, aluminum (Al) is oxidized to form Al3+ ions, while silver ions (Ag+) are reduced to form solid silver (Ag) on the electrode. The notation "Al(s) / Al3+(aq) // 3Ag+(aq) / 3 Ag(s)" indicates the reactants and products on each side of the cell, with the single vertical line representing the salt bridge or porous barrier between the two half-cells.

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36.

Diketahui E⁰ =Zn²⁺/Zn = -0,76 V Cu²⁺/Cu = +0,34 VFe²⁺/Fe = -0,41 V Ag⁺/Ag = +0,80 VPb²⁺/Pb = -0,13 VSel kimia yang berikut menghasilkan E⁰ sel yang paling besar adalah ….(E⁰ potensial sel)
- A.
  
  Zn/Zn2+(aq) // Cu2+(aq) /Cu(s)
- B.
  
  Fe/Fe2+(aq) // Ag+(aq) /Ag(s)
- C.
  
  Zn/Zn2+(aq) // Ag+(aq) /Ag(s)
- D.
  
  Pb/Pb2+(aq) // Cu2+(aq) /Cu(s)
- E.
  
  Zn/Zn2+(aq) // Pb2+(aq) /Pb(s)
Correct Answer
C. Zn/Zn2+(aq) // Ag+(aq) /Ag(s)

Explanation
The correct answer is Zn/Zn2+(aq) // Ag+(aq) /Ag(s). This is because the reaction between Zn and Ag+ has the highest E0 value (+0.80 V) compared to the other reactions. In electrochemistry, E0 represents the standard electrode potential, which measures the tendency of a species to gain or lose electrons. A higher E0 value indicates a stronger tendency for the species to gain electrons and be reduced. Therefore, the Zn/Zn2+(aq) // Ag+(aq) /Ag(s) reaction has the highest E0 value and will produce the largest cell potential.

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37.

Diketahui data potensial reduksi sebagai berikut:E⁰ Ca²⁺/Ca= -2,87 V E⁰ Pb²⁺/Pb= -0,13 VE⁰ Mg²⁺/Mg= -2,37 V E⁰ Au³⁺/Au= +1,50 VManakah reaksi yang diperkirakan dapat berlangsung?
- A.
  
  Mg(s) + Ca2+(aq) --> Mg2+(aq) + Ca(s)
- B.
  
  Pb(s) + Mg2+(aq) --> Pb2+(aq) + Mg(s)
- C.
  
  Pb2+(aq) + Au(s) --> Pb(s) + Au3+(aq)
- D.
  
  Ca(s) + Mg(s) --> Ca2+(aq) + Mg2+(aq)
- E.
  
  Mg(s) + Pb2+(aq) --> Mg2+(aq) + Pb(s)
Correct Answer
E. Mg(s) + Pb2+(aq) --> Mg2+(aq) + Pb(s)

Explanation
The given answer is correct because the reaction Mg(s) + Pb2+(aq) --> Mg2+(aq) + Pb(s) involves the reduction of Pb2+ to Pb, which has a more positive reduction potential (-0.13 V) compared to Mg2+ (-2.37 V). This means that Pb2+ is more likely to be reduced than Mg2+, making the reaction feasible. Additionally, Mg is oxidized to Mg2+, which has a more negative reduction potential, further supporting the feasibility of the reaction.

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38.

Diketahui potensial standar untuk reaksi sel berikut ini, Zn(s) +Fe²⁺(aq) → Zn²⁺(aq) +Fe(s) ; E_sel= 0,32VZn(s) +Ag⁺(aq) → Zn²⁺(aq) + Ag(s) ; E_sel=1,56VFe(s) + Ag⁺(aq) → Fe²⁺(aq) +Ag(s) ; E_sel=0,24VBerdasarkan harga-harga potensial di atas dapat disimpulkan bahwa urutan ketiga logam di atas dalam urutan reduktor yang menurun adalah….
- A.
  
  Ag, Fe, Zn
- B.
  
  Ag, Zn, Fe
- C.
  
  Fe, Zn, Ag
- D.
  
  Fe, Ag, Zn
- E.
  
  Zn, Fe, Ag
Correct Answer
E. Zn, Fe, Ag

Explanation
Based on the given standard potentials, the order of the metals in decreasing reducing power is Zn, Fe, Ag. This is because the metal with the more positive standard potential is a stronger reducing agent, meaning it is more likely to donate electrons and be oxidized in a redox reaction. In this case, Zn has the highest standard potential (0.32V), followed by Fe (0.24V), and Ag has the lowest standard potential (1.56V). Therefore, the correct answer is Zn, Fe, Ag.

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39.

Suatu sel volta menggunakan perak dan zink sebagai elektrode-elektrode :Ag⁺(aq) + e --> Ag(s) Eo = +0,80 VZn²⁺(aq) + e --> Zn(s) Eo = -0,76 VPernyataan yang benar adalah ….
- A.
  
  Ag bertindak sebagai anode
- B.
  
  Zn merupakan electrode positif
- C.
  
  Potensial sel sebesar 2,36 V
- D.
  
  Reaksi sel adalah 2Ag+(aq) + Zn(s) --> 2Ag(s) + Zn2+(aq)
- E.
  
  Ag mengalami oksidasi
Correct Answer
D. Reaksi sel adalah 2Ag+(aq) + Zn(s) --> 2Ag(s) + Zn2+(aq)

Explanation
The correct answer is "Reaksi sel adalah 2Ag+(aq) + Zn(s) --> 2Ag(s) + Zn2+(aq)". This is the correct balanced equation for the redox reaction that occurs in the voltaic cell. It shows the oxidation of Zn at the anode (Zn(s) --> Zn2+(aq) + 2e-) and the reduction of Ag+ at the cathode (2Ag+(aq) + 2e- --> 2Ag(s)). This balanced equation represents the overall reaction happening in the cell.

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40.

Di daerah industri, udara dapat mengandung gas-gas SO₂, CO₂, NO₂, O₂, N₂, dan CO₂. pasangan gas-gas yang dapat menyebabkan terjadinya korosi adalah….
- A.
  
  O2 dan N2
- B.
  
  CO dan NO
- C.
  
  CO2 dan CO
- D.
  
  SO2 dan NO2
- E.
  
  CO dan H2O
Correct Answer
D. SO2 dan NO2

Explanation
The correct answer is SO2 and NO2. In an industrial area, the air can contain gases such as SO2, CO2, NO2, O2, N2, and CO2. SO2 and NO2 gases are known to be corrosive. When these gases come into contact with certain materials, they can cause corrosion and damage to the surfaces. Therefore, the combination of SO2 and NO2 gases can lead to corrosion in industrial areas.

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41.

Apabila suatu logam mengalami korosi, logam tersebut ….
- A.
  
  Mengalami reaksi redukisi
- B.
  
  Mengalami reaksi oksidasi dan membentuk oksidanya
- C.
  
  Terhidrolisis membentuk larutan
- D.
  
  Menerima elektron dari O2 di udara
- E.
  
  Membentuk senyawa yang beracun
Correct Answer
B. Mengalami reaksi oksidasi dan membentuk oksidanya

Explanation
When a metal undergoes corrosion, it undergoes oxidation and forms its oxide.

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42.

Pada reaksi elektrolisa larutan NiSO₄ dengan elektroda Ag, reaksi yang terjadi pada anoda adalah....
- A.
  
  Ni2+ (aq) + 2e → Ni (s)
- B.
  
  Ag (aq) → ag+ (aq) + e
- C.
  
  Ni (s) → Ni2+ (aq) + 2e
- D.
  
  2H2O (l) + 2e → H2 (g) + 2OH− (aq)
- E.
  
  2H2O (l) → 4H+ (aq) + O2 (g) + 4e
Correct Answer
B. Ag (aq) → ag+ (aq) + e

Explanation
The correct answer is "Ag (aq) → ag+ (aq) + e". In the given electrolysis reaction of NiSO4 with Ag electrode, the reaction at the anode involves the oxidation of Ag atoms to Ag+ ions by losing one electron. This is represented by the equation Ag (aq) → ag+ (aq) + e.

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43.

Zat yang terbentuk di kutub negatif dari elektrolisis larutan NaCl dengan elektroda grafit adalah ....
- A.
  
  Logam Na
- B.
  
  Gas Cl2
- C.
  
  OH− dan H2
- D.
  
  H2O
- E.
  
  H+ dan O2
Correct Answer
C. OH− dan H2

Explanation
During the electrolysis of NaCl solution, the negative chloride ions (Cl-) migrate towards the positive electrode (anode) and undergo oxidation to form chlorine gas (Cl2). At the same time, water molecules (H2O) are also present in the solution. The positive hydrogen ions (H+) from the water molecules migrate towards the negative electrode (cathode) and undergo reduction to form hydrogen gas (H2). Additionally, hydroxide ions (OH-) are formed at the cathode due to the presence of water. Therefore, the correct answer is OH- and H2 as both are produced during the electrolysis of NaCl solution with a graphite electrode at the negative pole.

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44.

Perhatikan gambar elektrosis larutan KI dengan elektroda karbon. Zat yang terjadi pada elektrode B adalah...
- A.
  
  I2
- B.
  
  Gas H2
- C.
  
  Gas O2
- D.
  
  H2O
- E.
  
  Logam K
Correct Answer
A. I2

Explanation
Pada elektrode B terjadi reaksi reduksi, dimana ion iodida (I-) dari larutan KI menerima elektron dan berubah menjadi molekul iodin (I2). Oleh karena itu, zat yang terjadi pada elektrode B adalah I2.

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45.

Larutan ZnSO₄ dielektrolisis dengan arus listrik 5 ampere selama 10 menit. Bila Ar Zn = 65, endapan Zn yang terbentuk di katoda sebanyak....
- A.
  
  0,84 gram
- B.
  
  1,01 gram
- C.
  
  1,68 gram
- D.
  
  2,02 gram
- E.
  
  2,34 gram
Correct Answer
B. 1,01 gram

Explanation
During electrolysis, the amount of substance deposited on the cathode can be calculated using Faraday's law. The formula is:

Mass = (Current × Time × Atomic mass) / (Number of electrons × Faraday's constant)

In this case, the current is given as 5 Ampere, the time is given as 10 minutes, and the atomic mass of Zn is 65. The number of electrons involved in the reduction of Zn2+ to Zn is 2, and the Faraday's constant is approximately 96500 C/mol.

Plugging in these values into the formula, we get:

Mass = (5 A × 10 min × 65 g/mol) / (2 × 96500 C/mol)

Simplifying this equation gives us a mass of approximately 1.01 grams.

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46.

Proses elektrolisis lelehan NaCl dengan elektroda karbon, digunakan arus sebesar 10 ampere selama 30 menit. Massa logam natrium yang diperoleh adalah…..(Ar Na = 23, Cl = 35,5)
- A.
  
  A
- B.
  
  B
- C.
  
  C
- D.
  
  D
- E.
  
  E
Correct Answer
A. A

Explanation
The correct answer is A. In the process of electrolysis, the passage of an electric current through a molten compound or an aqueous solution causes a chemical reaction to occur. In this case, NaCl is being electrolyzed using carbon electrodes. The equation for the electrolysis of NaCl is 2NaCl -> 2Na + Cl2. From the equation, it can be seen that for every 2 moles of NaCl, 2 moles of Na are produced. Therefore, the mass of sodium obtained can be calculated using the formula: mass = moles x molar mass. Given that the current used is 10 amperes for 30 minutes, the number of moles of sodium can be calculated using Faraday's law. Finally, the mass of sodium obtained can be calculated using the molar mass of sodium.

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47.

Pada penyepuhan logam besi dengan tembaga digunakan larutan CuSO₄ dengan arus 2A selama 30 menit. Massa tembaga yang mengendap di katoda adalah...(Ar Cu = 63,5)
- A.
  
  (63,5 × 30 × 60) / (96.500)
- B.
  
  (63,5 × 96.500 × 2) / (2 × 30 × 60)
- C.
  
  (2 × 96.500 × 2) / (2 × 63,5)
- D.
  
  (63,5 × 2 × 30 × 2) / (60 × 96.500)
- E.
  
  (63,5 × 60 × 30 × 2) / (96.500)
Correct Answer
E. (63,5 × 60 × 30 × 2) / (96.500)

Explanation
The formula for calculating the mass of copper deposited on the cathode during the electroplating process is given by (63,5 × 60 × 30 × 2) / (96.500). This formula takes into account the atomic mass of copper (63,5), the time in minutes (30), the current in amperes (2), and the Faraday's constant (96.500). By plugging in these values into the formula, we can calculate the mass of copper deposited on the cathode.

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48.

Arus listrik sebesar I dialirkan ke dalam suatu sel ion Ag⁺ (Ar = 108) dan dalam waktu tertentu mengendapkan sebanyak 1,08 gram perak pada katoda. Jika jumlah listrik yang sama dialirkan melalui larutan yang mengandung ion X²⁺ (Ar = 40), maka jumlah logam X yang diendapkan pada katoda adalah....
- A.
  
  0,10 gram
- B.
  
  0,20 gram
- C.
  
  0,27 gram
- D.
  
  0,54 gram
- E.
  
  2,16 gram
Correct Answer
B. 0,20 gram

Explanation
The amount of metal deposited on the cathode is directly proportional to the amount of electricity passed through the solution. Since the same amount of electricity is passed through both the silver ion solution and the ion X2+ solution, the amount of metal deposited on the cathode in the ion X2+ solution will be proportional to the atomic mass of X (Ar = 40) compared to the atomic mass of silver (Ar = 108). The ratio of the atomic masses is 40/108, which simplifies to 5/27. Therefore, the amount of metal X deposited on the cathode will be 5/27 of the amount of silver deposited, which is 1.08 grams. Calculating this gives us 0.20 grams.

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49.

Sebanyak 1 liter larutan CrCl₃ 1M dielektrolisis dengan arus 6 Ampere. Jika diketahui Ar Cr = 52; 1 F = 96.500 maka waktu yang diperlukan untuk mengendapkan logam krom sebanyak 3,88 gram tersebut adalah....detik.
- A.
  
  A
- B.
  
  B
- C.
  
  C
- D.
  
  D
- E.
  
  E
Correct Answer
C. C

Explanation
The correct answer is C.

The question states that 1 liter of 1M CrCl3 solution is electrolyzed with a current of 6 Ampere. To determine the time required to deposit 3.88 grams of chromium metal, we need to use Faraday's law of electrolysis. According to Faraday's law, the amount of substance deposited is directly proportional to the charge passed through the solution. The charge can be calculated using the formula Q = I * t, where Q is the charge, I is the current, and t is the time. The molar mass of chromium (Cr) is 52 g/mol, so the moles of chromium deposited is 3.88 g / 52 g/mol = 0.0746 mol. Since the charge required to deposit 1 mole of chromium is 1 F (Faraday), the charge required to deposit 0.0746 mol of chromium is 0.0746 F. Thus, we can set up the equation 0.0746 F = 6 A * t. Solving for t, we get t = 0.0746 F / 6 A = 0.0124 s, which is equal to 12.4 milliseconds. Therefore, the time required to deposit 3.88 grams of chromium is 12.4 milliseconds.

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50.

Larutan 3 gram glukosa (Mr = 180) dalam 250 ml air, isotonik dengan larutan 4,6 gram X dalam 500 ml air. Zat X tersebut mempunyai massa molekul relatif ....
- A.
  
  69
- B.
  
  138
- C.
  
  276
- D.
  
  342
- E.
  
  684
Correct Answer
B. 138

Explanation
The question states that a solution of 3 grams of glucose (Mr = 180) in 250 ml of water is isotonic with a solution of 4.6 grams of X in 500 ml of water. To determine the relative molecular mass of X, we can use the concept of osmotic pressure. Osmotic pressure is directly proportional to the concentration of solute particles in a solution. Since the two solutions are isotonic, they have the same osmotic pressure, which means they have the same concentration of solute particles. By comparing the concentrations of glucose and X, we can calculate the relative molecular mass of X. The concentration of glucose is 3/180 = 0.0167 M, and the concentration of X is 4.6/500 = 0.0092 M. Since the concentrations are equal, we can set up the equation: 0.0167 = x/0.0092, where x is the relative molecular mass of X. Solving for x gives us x = 138, which is the correct answer.

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Kimia Trivia Quiz

Kemolalan larutan C2H5OH (Mr = 46) dengan persen berat 20% adalah ….

Tekanan uap larutan adalah ….

Fraksi mol urea (mr = 60) dalam larutan yang mengandung 15% urea adalah ….

Larutan H2SO4 0,4% (Mr = 98) dengan massa jenis 1,225 gram/mL mempunyai molaritas sebesar…

Volume larutan NaOH 0,25 M yang dibuat dengan melarutkan 1 gram NaOH adalah …. (Mr = 40)

Banyaknya air yang harus ditambahkan kedalam 200 mL larutan 0,4 M NaOH agar menjadi 0,1 M adalah…

Pernyataan yang tepat mengenai sifat koligatif larutan adalah ….

Nilai ΔTb dan ΔTf tergantung pada ….

Peristiwa berkurangnya tekanan uap larutan terjadi akibat ….

Besarnya penurunan tekanan uap larutan ….

Suatu zat non elektrolit (mr = 60) sebanyak 30 gram dilarutkan dalam 100 gram air (mr = 18) pada suhu 25oC. Bila tekanan uap jenuh pada suhu tersebut 23,76 mmHg, maka penurunan tekanan uap larutan tersebut adalah ….

Ke dalam 250 gram air ditambahkan 12 gram urea CO(NH2)2. Jika Kb air = 0,52oC, Kf air = 1,86°C, dan Ar C = 12, N = 14, O = 16, H = 1, maka titik didih larutan adalah ....

Tekanan uap jenuh air pada suhu 100oC adalah 76 cmHg. Tekanan uap larutan glukosa 10% massa (mr = 180) adalah .… mmHg.

Ke dalam 250 gram air ditambahkan 12 gram urea CO(NH2)2. Jika Kb air = 0,52oC, Kf air = 1,86°C, dan Ar C = 12, N = 14, O = 16, H = 1, maka titik beku larutan adalah ....

Kelarutan CaCl2 dalam air 0°C adalah sekitar 0,54 molal. Jika Kf = 1,86, maka penurunan titik beku larutan CaCl2 0,54 molal adalah ....

Pada suhu 27°C, glukosa (Mr = 180) sebanyak 7,2 gram dilarutkan dalam air sampai volume 400 mL. Jika R = 0,082 L atm/mol K, maka tekanan osmotik larutan yang terjadi adalah ....

Ke dalam 250 gram air dilarutkan 17,4 gram K2SO4 (Mr = 174). Jika Kb = 0,52°C/molal, maka kenaikan titik didih larutan tersebut adalah....

Larutan glukosa dalam air mendidih pada 100,26°C. Jika Kb = 0,52 dan Kf = 1,86 maka larutan tersebut akan membeku pada suhu ....

Diantara larutan 0,01 M di bawah ini yang memiliki tekanan osmotik yang paling besar adalah ....

Jika titik beku larutan glukosa 0,1 m dalam air adalah -0,18 °C, maka diharapkan titik beku 0,2 molal CaCl2 dalam air adalah ….

Sebanyak 3 g garam dapur (Mr = 58,5) dilarutkan dalam 1 kg air (Kf = 1,86) mempunyai titik beku -0,186oC. Derajat ionisasi garam dapur dalam air adalah ….

Larutan berikut yang isotonik dengan larutan NaCl 0,3 M adalah ….

Suatu larutan glukosa (Mr = 180) dalam 2 kg air (kb = 0,52) ternyata mendidih pada suhu 100,650C. Massa glukosa yang dilarutkan adalah ....

Suatu larutan terdiri atas 12 gram zat X (non elektrolit) di dalam 20 gram air. Jika larutan ini membeku pada -50 C dan penurunan titik beku molal air 1,860C, zat X tersebut mempunyai massa rumus relatif (Mr) ....

Data percobaan penurunan titik beku:LarutanKonsentrasi(molal)Titik Beku (°C)NaCl0,1–0,372NaCl0,2–0,744CO(NH2)20,1–0,186CO(NH2)20,2–0,372C6H12O60,1–0,186Berdasarkan data tersebut dapat disimpulkan bahwa penurunan titik beku bergantung pada ….

Bilangan oksidasi S pada ion S2O32- adalah ….

Di antara reaksi-reaksi tersebut di bawah ini yang merupakan contoh reaksi redoks adalah ….

Dalam reaksi MnO4- (aq) + C2O42-(aq) → Mn2+(aq) + CO2(g) suasana asam, persamaan C2O42- → CO2 melepaskan elektron sebanyak ….

Pada persamaan reaksi redoks:aMnO4- + 6H+ + bC2H2O4 → aMn2+ +8H2O + 10CO2 , maka a dan b berturut-turut adalah .…

Pada reaksi :aCr2O72-(aq) + 14H+ (aq) + bFe2+ (aq) --> cCr3+(aq) + 7H2O (l) + dFe3+(aq).Nilai a, b, c, dan d berturut-turut dalam reaksi setara adalah ….

Sel volta mengubah energi … menjadi energi ….

Diketahui potensial reduksi: Fe2+(aq) +2e- → Fe(s) E0 = -0,44VAl3+(aq) + 3e- → Al(s) E0= -1,66VMaka potensial sel untuk reaksi: 3Fe3+(aq) +2Al(s) → 3Fe(s) +2Al3+(aq) adalah ….

Suatu sel volta ditulis dalam diagram sel sebagai berikut :Cu(s) / Cu2+(aq) // 2Ag+(aq) / 2 Ag(s)Reaksi sel dari notasi di atas adalah ….

Reaksi sel dari suatu sel volta adalah ….Al(s) + 3Ag+ (aq) --> Al3+ (aq) + 3 Ag(s)Jika dari data tersebut ditulis dalam diagram sel adalah …..

Diketahui E0 =Zn2+/Zn = -0,76 V Cu2+/Cu = +0,34 VFe2+/Fe = -0,41 V Ag+/Ag = +0,80 VPb2+/Pb = -0,13 VSel kimia yang berikut menghasilkan E0 sel yang paling besar adalah ….(E0 potensial sel)

Diketahui data potensial reduksi sebagai berikut:E0 Ca2+/Ca= -2,87 V E0 Pb2+/Pb= -0,13 VE0 Mg2+/Mg= -2,37 V E0 Au3+/Au= +1,50 VManakah reaksi yang diperkirakan dapat berlangsung?

Suatu sel volta menggunakan perak dan zink sebagai elektrode-elektrode :Ag+(aq) + e --> Ag(s) Eo = +0,80 VZn2+(aq) + e --> Zn(s) Eo = -0,76 VPernyataan yang benar adalah ….

Di daerah industri, udara dapat mengandung gas-gas SO2, CO2, NO2, O2, N2, dan CO2. pasangan gas-gas yang dapat menyebabkan terjadinya korosi adalah….

Apabila suatu logam mengalami korosi, logam tersebut ….

Pada reaksi elektrolisa larutan NiSO4 dengan elektroda Ag, reaksi yang terjadi pada anoda adalah....

Zat yang terbentuk di kutub negatif dari elektrolisis larutan NaCl dengan elektroda grafit adalah ....

Perhatikan gambar elektrosis larutan KI dengan elektroda karbon. Zat yang terjadi pada elektrode B adalah...

Larutan ZnSO4 dielektrolisis dengan arus listrik 5 ampere selama 10 menit. Bila Ar Zn = 65, endapan Zn yang terbentuk di katoda sebanyak....

Proses elektrolisis lelehan NaCl dengan elektroda karbon, digunakan arus sebesar 10 ampere selama 30 menit. Massa logam natrium yang diperoleh adalah…..(Ar Na = 23, Cl = 35,5) ​

Pada penyepuhan logam besi dengan tembaga digunakan larutan CuSO4 dengan arus 2A selama 30 menit. Massa tembaga yang mengendap di katoda adalah...(Ar Cu = 63,5)

Arus listrik sebesar I dialirkan ke dalam suatu sel ion Ag+ (Ar = 108) dan dalam waktu tertentu mengendapkan sebanyak 1,08 gram perak pada katoda. Jika jumlah listrik yang sama dialirkan melalui larutan yang mengandung ion X2+ (Ar = 40), maka jumlah logam X yang diendapkan pada katoda adalah....

Sebanyak 1 liter larutan CrCl3 1M dielektrolisis dengan arus 6 Ampere. Jika diketahui Ar Cr = 52; 1 F = 96.500 maka waktu yang diperlukan untuk mengendapkan logam krom sebanyak 3,88 gram tersebut adalah....detik.

Larutan 3 gram glukosa (Mr = 180) dalam 250 ml air, isotonik dengan larutan 4,6 gram X dalam 500 ml air. Zat X tersebut mempunyai massa molekul relatif ....

Kemolalan larutan C₂H₅OH (Mr = 46) dengan persen berat 20% adalah ….

Larutan H₂SO₄ 0,4% (Mr = 98) dengan massa jenis 1,225 gram/mL mempunyai molaritas sebesar…

Nilai ΔT_b dan ΔT_f tergantung pada ….

Suatu zat non elektrolit (mr = 60) sebanyak 30 gram dilarutkan dalam 100 gram air (mr = 18) pada suhu 25^oC. Bila tekanan uap jenuh pada suhu tersebut 23,76 mmHg, maka penurunan tekanan uap larutan tersebut adalah ….

Ke dalam 250 gram air ditambahkan 12 gram urea CO(NH₂)₂. Jika K_b air = 0,52^oC, K_f air = 1,86°C, dan Ar C = 12, N = 14, O = 16, H = 1, maka titik didih larutan adalah ....

Tekanan uap jenuh air pada suhu 100^oC adalah 76 cmHg. Tekanan uap larutan glukosa 10% massa (mr = 180) adalah .… mmHg.

Ke dalam 250 gram air ditambahkan 12 gram urea CO(NH₂)₂. Jika K_b air = 0,52^oC, K_f air = 1,86°C, dan Ar C = 12, N = 14, O = 16, H = 1, maka titik beku larutan adalah ....

Kelarutan CaCl₂ dalam air 0°C adalah sekitar 0,54 molal. Jika K_f = 1,86, maka penurunan titik beku larutan CaCl₂ 0,54 molal adalah ....

Ke dalam 250 gram air dilarutkan 17,4 gram K₂SO₄(Mr = 174). Jika K_b = 0,52°C/molal, maka kenaikan titik didih larutan tersebut adalah....

Larutan glukosa dalam air mendidih pada 100,26°C. Jika K_b = 0,52 dan K_f = 1,86 maka larutan tersebut akan membeku pada suhu ....

Jika titik beku larutan glukosa 0,1 m dalam air adalah -0,18 °C, maka diharapkan titik beku 0,2 molal CaCl₂ dalam air adalah ….

Sebanyak 3 g garam dapur (M_r= 58,5) dilarutkan dalam 1 kg air (Kf = 1,86) mempunyai titik beku -0,186^oC. Derajat ionisasi garam dapur dalam air adalah ….

Suatu larutan glukosa (Mr = 180) dalam 2 kg air (kb = 0,52) ternyata mendidih pada suhu 100,65⁰C. Massa glukosa yang dilarutkan adalah ....

Suatu larutan terdiri atas 12 gram zat X (non elektrolit) di dalam 20 gram air. Jika larutan ini membeku pada -5⁰C dan penurunan titik beku molal air 1,86⁰C, zat X tersebut mempunyai massa rumus relatif (Mr) ....

Data percobaan penurunan titik beku:LarutanKonsentrasi(molal)Titik Beku (°C)NaCl0,1–0,372NaCl0,2–0,744CO(NH₂)₂0,1–0,186CO(NH₂)₂0,2–0,372C₆H₁₂O₆0,1–0,186Berdasarkan data tersebut dapat disimpulkan bahwa penurunan titik beku bergantung pada ….

Bilangan oksidasi S pada ion S₂O₃^2- adalah ….

Dalam reaksi MnO₄^- (aq) + C₂O₄^2-(aq) → Mn²⁺(aq) + CO₂(g) suasana asam, persamaan C₂O₄^2- → CO₂ melepaskan elektron sebanyak ….

Pada persamaan reaksi redoks:aMnO₄^- + 6H⁺ + bC₂H₂O₄ → aMn²⁺ +8H₂O + 10CO₂ , maka a dan b berturut-turut adalah .…

Pada reaksi :aCr₂O₇^2-(aq) + 14H⁺ (aq) + bFe²⁺ (aq) --> cCr³⁺(aq) + 7H₂O (l) + dFe³⁺(aq).Nilai a, b, c, dan d berturut-turut dalam reaksi setara adalah ….

Diketahui potensial reduksi: Fe²⁺(aq) +2e- → Fe(s) E⁰ = -0,44VAl³⁺(aq) + 3e- → Al(s) E⁰= -1,66VMaka potensial sel untuk reaksi: 3Fe³⁺(aq) +2Al(s) → 3Fe(s) +2Al³⁺(aq) adalah ….

Suatu sel volta ditulis dalam diagram sel sebagai berikut :Cu(s) / Cu²⁺(aq) // 2Ag⁺(aq) / 2 Ag(s)Reaksi sel dari notasi di atas adalah ….

Reaksi sel dari suatu sel volta adalah ….Al(s) + 3Ag⁺ (aq) --> Al³⁺ (aq) + 3 Ag(s)Jika dari data tersebut ditulis dalam diagram sel adalah …..

Diketahui E⁰ =Zn²⁺/Zn = -0,76 V Cu²⁺/Cu = +0,34 VFe²⁺/Fe = -0,41 V Ag⁺/Ag = +0,80 VPb²⁺/Pb = -0,13 VSel kimia yang berikut menghasilkan E⁰ sel yang paling besar adalah ….(E⁰ potensial sel)

Diketahui data potensial reduksi sebagai berikut:E⁰ Ca²⁺/Ca= -2,87 V E⁰ Pb²⁺/Pb= -0,13 VE⁰ Mg²⁺/Mg= -2,37 V E⁰ Au³⁺/Au= +1,50 VManakah reaksi yang diperkirakan dapat berlangsung?

Suatu sel volta menggunakan perak dan zink sebagai elektrode-elektrode :Ag⁺(aq) + e --> Ag(s) Eo = +0,80 VZn²⁺(aq) + e --> Zn(s) Eo = -0,76 VPernyataan yang benar adalah ….

Di daerah industri, udara dapat mengandung gas-gas SO₂, CO₂, NO₂, O₂, N₂, dan CO₂. pasangan gas-gas yang dapat menyebabkan terjadinya korosi adalah….

Pada reaksi elektrolisa larutan NiSO₄ dengan elektroda Ag, reaksi yang terjadi pada anoda adalah....

Larutan ZnSO₄ dielektrolisis dengan arus listrik 5 ampere selama 10 menit. Bila Ar Zn = 65, endapan Zn yang terbentuk di katoda sebanyak....

Proses elektrolisis lelehan NaCl dengan elektroda karbon, digunakan arus sebesar 10 ampere selama 30 menit. Massa logam natrium yang diperoleh adalah…..(Ar Na = 23, Cl = 35,5)

Pada penyepuhan logam besi dengan tembaga digunakan larutan CuSO₄ dengan arus 2A selama 30 menit. Massa tembaga yang mengendap di katoda adalah...(Ar Cu = 63,5)

Sebanyak 1 liter larutan CrCl₃ 1M dielektrolisis dengan arus 6 Ampere. Jika diketahui Ar Cr = 52; 1 F = 96.500 maka waktu yang diperlukan untuk mengendapkan logam krom sebanyak 3,88 gram tersebut adalah....detik.